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2

The Cayley-Hamilton theorem says that if $p$ is the characteristic polynomial of a matrix $A$, then $p(A) = 0$. Because $A$ commutes with itself and $I$, you can factor this matrix polynomial just like its scalar equivalent. In 2 dimensions, you are just reproducing this fact, since you are including both roots. In 3 dimensions, however, you have been trying ...


1

V isn’t just any arbitrary matrix. They're trying to take you step by step through the diagonalization process in this exercise, so the eigenvectors that you computed in part b are going to come into play here.


1

You started with a system of linear equations. In matrix form this can be written as: $$\left(\begin{array}{cc|c} 0 & -1 & 0 \\ 0 & -1 & 0 \end{array}\right)$$ from which you have gotten to $$\left(\begin{array}{cc|c} 0 & -1 & 0 \\ 0 & 0 & 0 \end{array}\right).$$ As we have a complete row of $0$'s, there might be infinitely ...


1

From your equations, you have $y=0$ and $x$ does not feature in the equations, so can be any value.


3

You have stated that $\{ v_1,v_2,\cdots,v_n\}$ is a basis of $V$ consisting of eigenvectors of $T$ with distinct eigenvalues $\{\lambda_1,\lambda_2,\cdots,\lambda_n\}$. And you have assumed that $S$ is another linear operator on $V$ that commutes with $T$. Because $TS=ST$, then $$ TSv_j = STv_j = \lambda_j Sv_j. $$ Because $\{ v_1,v_2,\cdots,v_n ...


0

We know that a rotation matrix has real entries: $${ \bf R_\theta } = \left[\begin{array}{rr} \cos(\theta) &-\sin(\theta)\\\sin(\theta)&\cos(\theta)\end{array}\right] = /a = \cos(\theta), b = \sin(\theta)/ = \left[\begin{array}{rr} a &-b\\b&a\end{array}\right]$$ Now consider $$|{\bf R_\theta} - \lambda {\bf I}| = (a-\lambda)^2+b^2 = 0$$ ...


0

Suppose $e^{i\theta}\neq e^{-i\theta}$. I leave to you the case $e^{i\theta}=e^{-i\theta}$. Let $v=v_1+iv_2$ be a column vector of $\mathbb{C}^3$ such that $Rv=e^{i\theta}v$, where $v_1,v_2\in\mathbb{R^3}$ (Prove that $v_1\neq 0$ and $v_2\neq 0$). Let $v^t$ be its transpose. Now, $R^tR=Id$ then $v=R^tRv=e^{i\theta}R^tv$. So $R^tv=e^{-i\theta}$. Next, ...


0

Thank you Augustin and Bryon. It seems that: There can in fact be many eigen vectors for eigenvalue 1 which are not probability vectors. There can also be many probability vectors (aka stationary probability vectors), for the eigenvalue 1. thank you for your answers.


4

$Sv_i$ is an eigenvector for $T$ with eigenvalue $\lambda_i$. The dimension of each eigenspace of $T$ is 1 since the eigenvalues are distinct, so $Sv_i = c_iv_i$ for some $c_i$.


0

It turns out that this is a well know property of the Hadamard product. See Horn and Johnson, doi: 10.1017/CBO9780511840371, Chapter 5. The proof is given by Elizabeth Million, Theorem 2.1 found at: http://buzzard.ups.edu/courses/2007spring/projects/million-paper.pdf


0

Recall that if $f(a)=0$ then there is a $q(x)$ that $f(x)=(x-a)q(x)$. in your question $-5^3+5^2+16.5+20=0$.


3

Note that $x^3-x^2-16x-20=(x-5)(x^2+4x+4)=(x-5)(x+2)^2$.


0

What you have done is correct (assuming the $x$s in your final answer should be $\lambda$s). But two points. I would say there is no need to avoid negatives in describing $F_7$, so your answer can be simplified to $-\lambda^3-\lambda^2+3\lambda-1$. (However your instructor may disagree - better check.) IMHO it would actually be easier if you used $F_7$ ...


0

A good way to think about eigenvalues/eigenvalues are that the eigenvectors are the vectors that are stretched or flipped(which you can think of "negative stretching"), and the eigenvalues are how much they get stretched by. So if we have some eigenvectors, say with eigenvalues 1,2,3. Then we have eigenvectors $v_1,v_2,v_3$. Let's say that $v_3$ is a ...


1

Finally, uranix hint with preconditioning lead me to a solution. The key performance problem comes from solving lots of systems of the form $Ax = b$ with our $A$. Fortunately, $A$ has so many nice properties, that the PCG algorithm works well when using ichol as a preconditioner. Thus using eigs' capability to take $x\mapsto A^{-1}x$ as a function leads to ...


2

Let $$ C=\pmatrix{I&0\\0&D}, $$ where the identity $I$ and an arbitrary symmetric $D$ have the same dimension greater than one. For $v_A=[1,0]^T$ and any nonzero $v_B$ (of the same dimension as $I$ and $D$), $v=v_A\otimes v_B$ is an eigenvector of $C$. The matrix $C$ does not need to have a Kronecker product form since $D$ is arbitrary symmetric.


2

This is in general false. Consider $v_A = v_B = (1, 1)^t$ and $C = \begin{pmatrix}1 & 2 & 3 & 4 \\ 2 & 1 & 4 & 3 \\ 3 & 4 & 1 & 2 \\ 4 & 3 & 2 & 1\end{pmatrix}$. Then $v_A \otimes v_B = (1, 1, 1, 1)^t$ and $C (v_A \otimes v_B) = 10 (v_A \otimes v_B)$, but clearly $C$ is not of the form $A \otimes I_2$. In ...


0

To help add some important concepts to eigenvalues and eigenvectors I will drag in another matrix. $${\bf A} = \left[\begin{array}{cc}2&2\\0&2\end{array}\right]$$ if we solve $$|{\bf A}-\lambda {\bf I}| = 0 \Leftrightarrow (2-\lambda)^2 = 0$$, we see that 2 is an eigenvalue that occurs twice. But if we solve $${\bf A}-2 {\bf I} = 0 ...


1

An eigenbasis is a basis in which every vector is an eigenvector. In your case, $$ \left\{ \pmatrix{-1\\1\\0}, \pmatrix{-1\\0\\1}, \pmatrix{1\\1\\1} \right\} $$ is an eigenbasis for your matrix $A$.


0

As stated in the comments: you've calculated the determinant incorrectly. When you fix your mistake, you should end up with the characteristic polynomial $$ \lambda^3 - 12 \lambda^2 + 36\lambda - 32 = (\lambda - 8)(\lambda - 2)^2 $$


2

The eigenvalue algorithm produces a diagonal matrix containing the eigenvalues of $A$. That matrix is positive definite if and only if the eigenvalues of $A$ are all positive. So, generally we will not get a positive definite eigenvalue matrix.


2

You can easily see that $$A^nx^{(0)} = \sum_{i=1}^N\alpha_i \lambda_i^n x_i$$ You can rewrite that as $$\lambda_1^n\sum_{i=1}^N \alpha_i\left(\frac{\lambda_i}{\lambda_1}\right)^nx_i$$ and then study what happens to $\frac{\lambda_i}{\lambda_1}$ if $i=1$ or of $i\neq 1$.


2

Since $A$ is real symmetric, there exists a matrix $P$ such that $$PAP^{-1} = D =\verb/diag/(\lambda_1,\lambda_2,\ldots,\lambda_n)$$ is diagonal with eigenvalues $\lambda_1,\ldots,\lambda_n$. This leads to $$P (I_n+A)^{-2}P^{-1} = (I_n + PAP^{-1})^{-2} = \verb/diag/\left(\frac{1}{(1+\lambda_1)^2},\ldots,\frac{1}{(1+\lambda_n)^2}\right)\\ \implies ...


4

No. The eigenvalues of $A$ are the zeroes of the characteristic polynomial. It doesn't matter whether you take the characteristic polynomial to be $\det(A -\lambda I)$ or $\det(\lambda I - A)$; they differ by at most a factor of $-1$, so their zeroes are the same.


0

The name "min-max theorem" given in the comments lead to the following very elegant proof. To repeat in our context: given a symmetric matrix A, we construct an orthonormal basis $\{ v_1,b_2,...,b_n \}$ to $\mathbb{R}^n$ using the eigenvectors of A (which is possible due to the symmetry of A), using the eigenvector $v_1$ from the theorem. As we only ...


1

Your questions regarding the usefulness of the index and finding a Jordan basis I will answer together: as with diagonalization one usually finds the Jordan form first and then determine the basis of vectors that transforms the given matrix to Jordan form. One "useful" way to determine the Jordan form of a matrix $A$ is to find the Segre characteristics, ...


1

We can always do so if $D$ is diagonalizable invertible. Otherwise, we have counterexamples like $$ D=\pmatrix{0&1\\0&0} $$ Note that $D^2$ as the eigenvector $(0,1)$.


2

Hint: In a basis of eigenvectors, the matrix of the endomorphism is diagonal, and the correspondence between the matrix $A$ in the original basis and the matrix $D$ in the basis of eigenvectors is given by: $$D=P^{-1}AP.$$


2

By Spectral Theorem, $A$ is orthogonally similar to a diagonal matrix, i.e $$ P^{T}AP=\pmatrix{\lambda_1 \\ & \ddots \\ && \lambda_n} $$ where $\lambda_i$ is eigenvalue of $A$, and $\space P^{T}P=P^{-1}P=I$. Since $$ \pmatrix{\lambda_1 \\ & \ddots \\ && \lambda_n}=Q\pmatrix{I_r\\ &-I_s\\ && 0}Q $$ where $r$ is number of ...


0

The Jordan form is diagonal if the matrix has a basis of eigenvectors. If you have a basis $\{ x_1,x_2,\cdots x_m \}$ of $E(\lambda_1)$ and a basis $\{ y_1,y_2,\cdots,y_{n-m}\}$ of $E(\lambda_2)$, then the union of these two bases is a basis of the full space because, if $$ (\alpha_1 y_1 + \alpha_2 y_2 + ...


2

Note that the trace of any permutation is its number of fixed points (that is, the number of $1$-cycles). As in my previous answer, let $x_{ij}$ denote $x_ix_j = x_{ji}$. In order to have $A^{(2)}(x_{ij}) = x_{ij}$, we must have $$ (Ax_i)(Ax_j) = x_{ij} \DeclareMathOperator{\tr}{Tr} $$ There are precisely two ways in which this can occur: $x_i$ and $x_j$ ...


1

If $A$ acts on this basis than the trace is simply going to be the size $N$ of the set of basis vectors that it fixes (i.e. the number of $i$ such that $A(x_i) = x_i$) and as such we have that $Tr(A) = \sum_{i = 1}^n \lambda_i = N$. Then the number of basis vectors fixed by $A^{(2)}$ is simply $\frac{\text{Tr}(A^2) + \text{Tr}(A)^2}{2} - N$ which is ...


3

Hint: If $\lambda$ is an eigenvalue of $A$, let $x$ be the associated eigenvector, and consider $x'Ax$.


2

Suppose our matrix $A$ has eigenvalue $\lambda$. If $\lambda = 0$, then there is some eigenvector $x$ so that $Ax = 0$. But then $x^T A x = 0$, and so $A$ is not positive definite. If $\lambda < 0$, then there is some eigenvector $x$ so that $Ax = \lambda x$. But then $x^T A x = \lambda \lvert x \rvert^2$, which is negative since $\lvert x \rvert^2 ...


1

Hint: Since all the columns are the same, the matrix has rank $1$. Thus, all but one of the eigenvalues are $0$. Can you guess an eigenvector which corresponds to the non-zero eigenvalue, and determine this non-zero eigenvalue?


0

This is false, it does not hold disregarding wether the matrix is diagonalizable or not. For a non diagonalizable matrix, take $A=[-2,1;0, -2]$. For a diagonalizable matrix, take $A=[-1,0; 3, -0.2]$. Edit: the textbook I mencioned in the question was incomplete. The correct theorem is that there exist a basis for which this holds. For source, go to ...


0

That's not actually a good basis to work with. If your vectors are $\vec v_1, \ldots, \vec v_k$, then there should be distinct integers $n_1,\ldots,n_k$ such that: The $n_k$'th entry of $\vec v_k$ is $1$. The $i$'th entry of $\vec v_k$ is $0$, for $i>n_k$. The $n_k$'th entry of $\vec v_j$ is $0$, for $j\not=k$. So I will use the basis ...


0

It sounds from the context like you're looking for a square matrix. One way to get a matrix with the right nullspace is to work backwards. That is, if you used row-reduction (Gaussian elimination) to get the matrix into a reduced form, what would it have to look like in order to have this null-space? In fact, the only row-reduced $3 \times 3$ matrix with ...


2

When you talk about the eigenvalues of $A^{(2)}$ over $V^{(2)}$, you assume that the image of $A^{(2)}$ lies within $V^{(2)}$ (or, in other words, that $V^{(2)}$ is an invariant subspace of $A \otimes A$). Note, however, that this is not generally the case. For example, we can consider the map $A$ over $\langle x_1,x_2 \rangle$ defined by $$ Ax_1 = Ax_2 = ...


6

Note that $(A - I)^2 = A^2 - 2A + I = 0; \tag{1}$ thus for all $n$-vectors $x$, $(A - I)^2 x = 0; \tag{2}$ if $(A - I)x = 0 \tag{3}$ for all $x$, then $Ax = Ix = x \tag{4}$ for all $x$ as well; in this case, $A = I$ is the identity matrix, and every nonzero vector is an eigenvector of $A$ with eigenvalue $1$. If, on the other hand, there is some $x$ ...


4

Let $x$ be an eigenvector of $A$ so that $Ax=\lambda x$, then $$(A^2-2A+I)x=A(Ax)-2Ax+Ix=\lambda Ax-2\lambda x +x=(\lambda^2-2\lambda+1)x=0$$ And with $x\neq 0$ we have $\lambda=1$ As per comments below we need to know we have an eigenvector in the first place. This pretty much gets straight back to Robert Israel's answer. We have $(A-I)^2=0$ so that ...


0

You have $(A-I)^2=0$, hence $(x-1)^2$ is a divisible by the minimal polynomial of $A$, which is therefore $x-1$ or $(x-1)^2$. In both cases, as the eigenvalues are the roots of the minimal polynomial, the only eigenvalue is $1$. In the first case, $A =I$; in the second case, the matrix is not diagonalisable.


1

You should notice the factorization as a square as if $A$ were a real number: $$ A^2-2A+I=(A-I)^2=\bf{0} $$ where $\bf{0}$ denotes the zero matrix of dimension $n\times n$. If $A-I$ were injective, the zero map would have to be injective, but clearly it is not.


6

Hint: For any nonzero vector $x$, $(A - I)^2 x = 0$. If $(A-I) x = 0$ then ... If not, then ...


2

This means that the minimal polynomial of $A$ divides $x^{2}-2x+1=(x-1)^{2}$ and hence is either $(x-1)$ or $(x-1)^{2}$. So $1$ is a root of the minimal polynomial of $A$ hence an eigenvalue of $1$.


3

It is not true, I think you mean diagonal matrix instead of diagonalizable. As an example you can take the 2x2 matrix $A$=$\begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix}$, whose eigenvalues are $3$ and $-1$. If you meant diagonal then you can check it almost from the definition. Another way to check it is thinking that a matrix must have the same ...


0

Notice that $Z$ does not need to be diagonal. Thus, $Z_{ii}^2$ need not be an eigenvalue of $X$. In fact $A$ and $X$ are simultaneous diagonalizable if and only if $AX = XA$. Now the proof: We have $$A - X = U D^2 U' - U Z^2 U' = U (D^2 - Z^2) U' \succeq 0$$ if and only if $$ D^2 - Z^2 \succeq 0 $$ as $U$ is unitary. Now, let $e_i$ be the unit vector ...


1

The zero operator has a nice basis of eigenvectors, all with the same eigenvalue.


0

It's not completely clear what method MAGMA uses by default for finite fields, but it looks like the Hessenberg algorithm. It seems like once the matrix is reduced to this form, it is easy to find the eigenvalues/eigenvectors, and the eigenvectors can be transformed back to eigenvectors of $A$ with the saved similarity matrix. It has a few options though as ...


1

$det(A-\lambda I)=(\lambda+3)^2(\lambda-1)(\lambda-5)$. Hence $det(A-\lambda $I) not zero except $\lambda \neq-3,1,5$



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