Questions tagged [diagonalization]
For questions about matrix diagonalization. Diagonalization is the process of finding a corresponding diagonal matrix for a diagonalizable matrix or linear map. This tag is NOT for diagonalization arguments common to logic and set theory.
2,584 questions
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Why is a real symmetric matrix diagonalizable?
Exactly, I can't understand the real symmetric matrix is diagonalizable only from the symmetry.
I can prove that the diagonalization of this kind of matrices by mathematical induction,as in Artin's ...
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Can one construct a liar's sentence using the diagonal lemma and the principle of explosion?
Isn't it possible to form the liar's sentence by using the diagonal lemma to form the sentence "This sentence proves all statements"?
I.e. For some classical first order theory $T$ ...
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Inductive proof of necessary and sufficient conditions for diagonalisability
I am very confused by the following highlighted lines in a proof of necessary and sufficient conditions for diagonalisability in Linear Algebra Done Right (4th ed.), Axler S. (2024).
Questions.
How ...
- 1,106
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35
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Definite unimodular form equivalent to a diagonal form while itself being non-diagonalizable
In a lecture on exotic $\mathbb R^4$s, Robert Gompf claims that the following bilinear forms are "equivalent" (I presume, over the integers):
$$
\left(\begin{array}{c|c}
1 & 0\\\hline
0 &...
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votes
2
answers
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Showing that two $2\times 2$ matrices are similar
I want to prove that the following $2\times 2$ matrices are similar over $\mathbb R$:
$$
A=\begin{pmatrix}
0 & -1\\
1 & 0\\
\end{pmatrix}
\qquad
B=\begin{pmatrix}
-3 & -5\\
2 & 3\\
\...
- 5,715
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votes
1
answer
151
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Classify for which $M$ is $M+𝛼M^*$ diagonalizable for arbitrary $𝛼≠0$?
For which $M∈\text{Mat}(n×n, ℂ)$ is $M+𝛼M^*$ diagonalizable for arbitrary $𝛼∈ℂ \setminus \{0\}$?
Alternatively just choose $𝛼∈ℝ \setminus \{0\}$ and analyse, when $M+𝛼M^\mathrm{T}$ is ...
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Exponential of a Non-Triangular Matrix is Non-Triangular
I'd like to show that, given $A$ a non-upper triangular matrix, then $\exp(A)$ mustn't be upper triangular either. Equivalently, I could show that the logarithm of an upper triangular matrix is always ...
- 107
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3
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82
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For a given matrix with real parameter $\alpha$, determine if there is an $\alpha$ such that $\beta(x, y)=y^T A_\alpha x$ is an inner product.
I've recently practiced some old linear algebra exams and came across this question.
Given $\alpha \in \mathbb{R}$ and
$$
A_\alpha=\left[\begin{array}{lll}0 & 0 & \alpha \\ 0 & 1 & 0 \\...
0
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1
answer
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Diagonalization of an upper triangular matrix
I am considering a certain Lie algebra, in particular a complex upper-triangular Lie algerbra. Furthermore, I wish to find a nice way to write the exponential of an arbitrary element in this upper-...
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Is this diagonalization formula incorrect?
Note: After posting I realized that it may be difficult to distinguish between $\mathfrak{x}_{\bar{i}}$ and $\mathfrak{x}_{\tilde{i}}$ when viewed with a web browser. I suggest using ...
- 3,859
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1
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Efficiently computing the trace of products of diagonalizable matrices
Suppose that I have two matrices $A$ and $B$ which are both symmetric: $A=A^T, B=B^T$. Moreover, I know how to diagonalize both $A$ and $B$.
Now I would like to define $T=A^{1/2}BA^{1/2}$, which is ...
- 391
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answer
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Alternative attempt to prove self-adjoint spectral theorem - can this work?
I've read the standard proof of the self-adjoint spectral theorem which uses induction, but I thought of another idea and am attempting to make it work, I'd like to know if I'm on to something or if ...
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Given a matrix of the form $M=[(A,B),(0,I)]$, show $M$ is diagonalizable iff $A$ is diagonalizable
This question in the sample final for my linear algebra class has me, for the first time in this class, stumped (and the prof hasn't released a solution set):
Suppose $M$ is a matrix of the form $\...
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2
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Extending a pair of left- and right-eigenvectors to a full eigenbasis
When a matrix is diagonalizable, it is possible to find left and right eigenvectors $u_i$ and $v_j$ such that $u_i^T v_j = \delta_{ij}$ (Kronecker delta), $u_i^T A = \lambda_i u_i^T$, and $A v_i = \...
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Coprime annihlating polynomials
Only the zero matrix admits two coprime annihilating polynomials. Is this true or false.
I think it's true but the notebook I use claims it's false without elaborating.
My approach is the following:
...
- 1
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0
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Does diagonalization of a matrix imply that the eigenvectors are unique to a given form of a matrix? [duplicate]
If you have a collection of n (nonzero and unique) eigenvectors, is there a way to find a general form of an n-by-n matrix that corresponds to them in such a way that 'rules out' alternative forms?
...
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Smooth Diagonalization of a Symmetric Operator on a Manifold
I'm studying the shape operator of Hopf hypersurfaces, and in this setting, all of its eigenvalues are supposed to be constant. I'm wondering whether there is a theorem that guarantees the following: ...
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Geometric (or physical) interpretation of congruent transformation
It's a bit hard to construct this question precisely, I may do an analogy to similar transformation.
From the passive point of view, two matrix connected by a similar transformation can be viewed as ...
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Let $P$ be a symmetric matrix and $P\in M_{n} (\mathbb{F} )$. Is it similar to the following matrix block-diagonal form?
We know that for a symmetric matrix $P$ on a $n$-dimensional real linear space $V$, it always can be diagonalized. The reason lies in the proposition that
$V$ can be expressed as the sum of all ...
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If $U=V_{\lambda _1} \oplus\dots\oplus V_{\lambda _m}$and $W$ is an $A$-invariant subspace and $W + U = V$, then $U + W $ is not a direct sum.
The following discussions are all restricted to finite-dimensional linear spaces and eigenvalues are by default discussed in the field of complex numbers.
We know that for a symmetric transformation $...
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row-equivalence of a matrix always diagonalizable implies what?
This is an exercise given in an entrance exam to a highly regarded French engineering school:
Let $A$ a square matrix of size $n$ over a certain field (The real field or the complex field for instance)...
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1
answer
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Why does the golden ratio appear as a singular value of $\begin{pmatrix}1&1\\0&1 \end{pmatrix}$?
The matrix $\begin{pmatrix}1&1\\0&1 \end{pmatrix}$ is "special". For example, it is perhaps the first example a student comes across of a non-diagonalisable matrix. Moreover, since $\...
- 1,024
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3
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Let $n$ be a positive integer. Find $A^{4n}$ in terms of $n$.
I am presented with a matrix $A$
\begin{equation*}
A = \begin{bmatrix}
-8 & -30 & 10 \\
5 & 20 & -7 \\
10 & 33 & -10
\end{bmatrix}
\end{equation*}
and have been asked to find ...
- 481
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vote
0
answers
79
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Diagonalizing a positive semidefinite matrix-valued function
Consider a $C^1$ function $\Sigma:\mathbb{R}^d \to \mathbb{R}^{d\times d}$ such that $\Sigma(\theta)$ is positive semidefinite for all $\theta \in \mathbb{R}^d$. Is it possible to find matrix-valued ...
- 476
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What does it mean for an element of $\operatorname{SL}_2(q)$ to be semisimple but not diagonalisable? That is, what does it look like in general?
The Question:
What does it mean for an element of $\operatorname{SL}_2(q)$ to be semisimple but not diagonalisable? That is, what does it look like in general?
The Details:
Recall that $\...
- 48.5k
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If matrix $A$ has a left inverse $C$, is $AC$ orthogonally diagonalizable with all eigenvalues $0$ or $1$?
Let $A,C$ be matrices such that $CA = I$. Then $AC$ is defined. I conjecture the following:
$AC$ is orthogonally diagonalizable.
Its eigenvalues are either $1$ or $0$.
It is Hermitian
Informally, ...
- 6,461
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votes
2
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I have a system of $6$ matrix equations, where the matrices are $64 \times 64$. How can I approach this problem in a more efficient way?
I think this question is more of a mathematics question rather than a programming question, read below for further details.
Goal
I have a system of somwehat big ($64 \times 64$) matrix equations of ...
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Block Diagonalization of a $3\times3$ real matrix with complex eigenvalues
let $A$ be real matrix such the its characteristic polynomial has only one real eigenvalue $\lambda$ so, necessarily it has two conjugated complex eigen values $\beta$ and $\bar{\beta}$.
By the ...
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Diagonalizability of transpose (dual) operator for infinite-dimension vector spaces (purely algebraic version)
I'm writing some notes on Linear Algebra and thought of the following question:
What is the relation between diagonalizability of a linear operator and its dual.
Here are my definitions: A linear ...
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Artin-Wedderburn decomposition of equivariant matrices
Let $V$ be an $N$-dimensional real inner product space with an orthonormal basis $\beta$. Let $G$ act on $\beta$ by permuting its elements and identify $\text{End}_G(V)$ as a subspace of $\text{Mat}_N(...
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Real symmetric matrices whose eigenvectors don't have eigenvalue -1
I wonder if we can always parameterize the eigenvectors $V$ in an eigendecomposition of real symmetric matrices with Cayley's parameterization, under which any orthogonal matrix with −1 as an ...
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If $f$ is diagonalizable and its characteristic polynomial has simple roots then $g$ is diagonalizable.
Let $\mathbb{V}$ be a finite dimension vector space over a field $\mathbb{k}$ and let $f,g:\mathbb{V}\to\mathbb{V}$ be endomorphisms such that $f\circ g=g\circ f$. Prove that if $f$ is diagonalizable ...
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views
Let $f$ be an endomorphism such that $f+f^*=0$. Prove $f$ is diagonalizable [duplicate]
Let $\mathbb{V}$ be a finite-dimensional inner product vector space over $\mathbb{C}$ and $f:\mathbb{V} \to \mathbb{V}$ an endomorphism such that $f+f^*=0$ with $f^*$ being the adjoint function of $f$....
- 146
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votes
0
answers
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views
Trace of an orthogonal matrix not equal to sum of eigenvalues
I was taught that any orthogonal matrix must have eigenvalues of magnitude 1, i.e. $\lambda_i=\pm1,\forall i$. However, I am given the following matrix
$$
M = \frac{1}{2}\begin{pmatrix}
-1 & -1 &...
- 11
1
vote
1
answer
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views
Diagonalisation of a real square matrix when it is also symmetric.
A standard result in linear algebra is that for a real, square matrix $A \in \mathbb{R}^{n \times n}$ can be diagonalised as follows,
$$A = U \Lambda U^T = U \Lambda U^{-1}.$$
Where,
$\Lambda \in \...
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How to diagonalize Hermitian antilinear (conjugate linear) operators?
I am trying to find all eigenvectors and eigenvalues of a Hermitian antilinear operator because the problem appears in the computation of complex rational minimax approximations. Antilinear operators ...
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Can the complex spectral theorem be proven from complex SVD?
Using a little trickery, one can prove the real spectral theorem from the real SVD: Can the spectral theorem from linear algebra be proved easily using the SVD?
Can one do this for complex instead of ...
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Symplectic diagonalization of positive semi-definite matrices.
If $H$ is a positive definite matrix, it is well known by Williamson's theorem that one can brought $H$ into the block diagonal form $S^T H S = \begin{pmatrix} D & 0 \\ 0 & D \end{pmatrix}$ ...
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$A=\left(\begin{array}{cccccccccccccccccccc}a^2&ac&c^2\\ 2ab&ad+bc&2cd\\ b^2&bd&d^2\end{array}\right)$ is Diagonalizable
In $\Bbb C$, how to prove that $A=\left(\begin{array}{cccccccccccccccccccc}a^2&ac&c^2\\
2ab&ad+bc&2cd\\
b^2&bd&d^2\end{array}\right) $ is Diagonalizable if and only if $ \...
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answer
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If $A, B$ are $4$-by-$4$ diagonalizable matrices such that rank($A$)+rank($B$)=$5$, can $(AB)^2 = 0$?
If $A,B \in \mathbb{R}^{4\times4}$ are diagonalisable matrices such that $rank(A)+rank(B)=5$, is it possible $(AB)^2 = 0$? That is, can matrix $AB$ be nilpotent?
My work: I think it may be impossible. ...
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Diagonalizing each block of block diagonalized matrix
We have a $n\times n$ matrix $A$ , which is also block diagonalized into
$$
\left(\begin{matrix}
A_1 & & &\\
& A_2 & &\\
& & \ddots & \\
& & & A_s
\...
- 21
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vote
1
answer
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Simultaneously Matrix Diagonalization
Given $ A_1, \dots, A_M \in \mathbb{R}^{n \times n}$, I am wondering whether there exist matrices $X, Y \in \mathbb{R}^{r \times n} $ (with $ r $ potentially larger than $ n $) such that $ X A_k Y^T $ ...
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Matrix $B$ with entries in an integral domain $R$ is diagonalizable if $A$ is diagonalizable with distinct eigenvalues and $AB = BA$.
This is exercise 7.13 from Chapter 6 of Aluffi's Algebra: Chapter 0 textbook. It says that if $R$ is an integral domain, $A \in M_{n}(R)$ is diagonalizable with distinct eigenvalues and $AB = BA$ (...
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Prove that the operator is diagonalizable
I need help finding the source or an idea to solve the following problem:
Suppose that $V$ is a finite-dimensional inner product space over $\mathbb{C}$, and $L:V\to V$ is an operator. Assume that the ...
- 820
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votes
0
answers
38
views
Are JNF with identical blocks in different orders considered equal?
I'm working on a linear algebra problem involving JNF and similarity transformations. Here’s the setup: Given a matrix
M = \begin{pmatrix} 7 & 0 & 0 \\ 0 & 7 & 0 \\ -1 & 4 & 7 ...
- 17
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votes
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372
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If $T^2$ has an upper-triangular matrix then $T$ has.
Is this question true:
Prove or give a counterexample: If $T \in \mathcal{L}(V)$ and $T^2$ has an upper-triangular matrix with respect to some basis of $V,$ then $T$ has an upper-triangular matrix ...
- 3,153
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Is it possible to diagonalize the Hadamard product of mutually-diagonalizable PSD matrices?
I've searched around and it seems that little can be said in general about the spectrum of Hadamard (elementwise) products. Thankfully, my problem has structure: Given two mutually-diagonalizable PSD ...
- 111
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votes
0
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62
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Eigenvalues of the product of diagonalizable matrices
I am trying to understand how I can compute the eigenvalues of $A^{-1}B$ where $A$ and $B$ are diagonalizable matrices (in patricular, they are symmetric tridiagonal matrices).
I know that $A$ can be ...
0
votes
1
answer
68
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Null space of restriction and diagonalizability
I was practising with exercises on Jordan canonical form in Section 7.1 of Friedberg et al. Linear Algebra, in the point f) of exercise 7 it's said to prove that given a diagonalizable linear operator ...
- 65
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votes
1
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Invertible Matrix for diagonalisation for solving differential equations
For the following question I've gotten a matrix $P =\begin{pmatrix}
1 & 1 & 1\\
1 & 0 & 1\\
0 & 1 & 1
\end{pmatrix}$ that works for $P^{-1}AP=J$ but doesn't work for the ...
