# Tag Info

## New answers tagged eigenvalues-eigenvectors

1

You need a basis of the generalized eigenspace. Therefore you need to make sure that the vectors are linearly independent. You have $v_1,v_2$ and now you need to complete them to a basis of the generalized eigenspace by adding a vector from the eigenspace that is independent of the two. (It actually suffices that it is independent of $v_1$ but I find it ...

2

No. Consider an $n$-dimensional matrix $A$ that is $0$ except for $a_{11} = c \neq 0$. Then $c$ is an eigenvalue, yet $A - c I$ has rank $n-1$ while the rank of $A$ is $1$. More generally the rank of $A-cI$ is $n$ minus the (geometric) multiplicity of the eigenvalue and at least $n$ minus the arithmetic multiplicity of the eigenvalue that is the ...

0

The Hilbert spaces of Quantum Mechanics must be separable. That's basically an axiom of QM that is a requirement of constructibility. That means that a selfadjoint operator can never have more than a countable number of eigenvectors. For example, $$L=\frac{1}{i}\frac{d}{dx} \\ \mathcal{D}(L)=\{ f \in \mathcal{AC}\subset L^2[0,2\pi] : f'\in ... 1 For a block matrix$$M = \pmatrix{U & V\cr W & X\cr}$$where U, V, W, X are square matrices and W and X commute, \det(M) = \det(UX - VW). In this case, for \lambda \ne 0 that says$$ \eqalign{\det(C - \lambda I) & = \det(\lambda^2 I - (\lambda + 1) AB) =\lambda^{4m} \det(I - \lambda^{-2}(\lambda+1) AB)\cr \det(D - \lambda I) & ...

1

Assume that $m\geq n$. We know that $\det(xI-AB)=x^{m-n}\det(xI-BA)$. Assume that $\lambda\not= -1$. One has $\det(C-\lambda I)=\det(\lambda^2I-(\lambda+1)AB)=(\lambda+1)^m\det(\dfrac{\lambda^2}{\lambda+1}I-AB)$ and, in the same way, $\det(D-\lambda I)=(\lambda+1)^n\det(\dfrac{\lambda^2}{\lambda+1}I-BA)$. Putting $x=\dfrac{\lambda^2}{\lambda+1}$, one ...

2

Separation of variables works on regions that are rectangular in some particular coordinate system. The underlying operator must be separable in that coordinate system. For the Laplacian, that generally means an orthogonal coordinate system such as spherical coordinates, cylindrical coordinates, elliptic coordinates, etc.. A rectangular region in spherical ...

0

There are several things to be said here, but first: a similar statement cannot hold for anti-commuting matrices. For example, take $$A=\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}, B=\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}.$$ Then $A$ and $B$ anti-commute and they both have $1$ and $-1$ for eigenvalues. Take $P(x,y)=xy$. Then ...

2

Actually, we have that $\operatorname{ker}p(A)$ is $A$-invariant for any polnomial $p \in \mathbb C[x]$. The proof goes like this: Notice that $A$ commutes with any polnomial expression of $A$, i.e. $Ap(A)=p(A)A$. Now, let $v \in \operatorname{ker}p(A)$. We have $p(A)Av=Ap(A)v=0$, hence $Av \in \operatorname{ker}p(A)$, which is the desired result.

1

Let $E_{\lambda,j}=\text{ker} (A-\lambda I)^j$ and $y\in E_{\lambda,j}$. You have to show that $Ay\in E_{\lambda,j}$, i.e. that $(A-\lambda I)^jAy=0$. Just use the fact that $y\in E_{\lambda,j}$ and that $(A-\lambda I)^jA=A(A-\lambda I)^j$ (this last point should be justified).

0

In the case in which $V$ has finite dimension, such an operator has $n$ eigenvalues, where $n$ is the dimension of $V$. Eigenvalues are the roots of the characteristic equation, that is a polynomial equation over F. Such roots cannot be more than n (but can be less) in F itself. In this case they are n and are all equal. See it also from a geometrical point ...

1

You are given that $A^2(A^2-4)=0$. Now If $A$ is invertible, multiply by $A^{-2}$ to get $A^2-4=0$; if in addition it is given that $A$ is not a scalar matrix, its minimal polynomial has degree${}>1$, so must be $X^2-4$. Here we remark that $X^2(X^2-4)=X^2(X+2)(X-2)$ splits over$~\Bbb R$. If $A$ were annihilated by a spilt polynomial without repeated ...

0

First attempt: $C$ has the block structure $$C = \pmatrix{0&I\\A&B}$$ In particular, following the discussion here, we note that since the top two block commute, we have $$\det(C - \lambda I) = \det \pmatrix{-\lambda I&I\\A&(B - \lambda I)} = \\ \det[(-\lambda I)(B - \lambda I) - (I)(A)]=\\ \det[\lambda^2 I - \lambda B - A]$$ Second ...

0

So the added constraint to your problem is that eigenvalues $a$ and $b$ must be mapped to $-1$ and $1$. One possible solution is what follows. If T is the diagonalizing matrix of M, then $M=TDT^{-1}$, where $D=\begin{bmatrix} a&0&0&\dots\\ 0&b&0&\dots\\ 0&0&c&\dots\\ \vdots&\vdots&\vdots&\ddots \end{bmatrix} ... 0 I found out that the following transformation works, $$Ms=\frac{M−((b+a)/2) I}{(b−a)/2}$$ where I is a Identity matrix.$M_s$also has the same eigenvectors as the original matrix$M$. 0$M_S=sign(a+b)/max(|a|, |b|) M$0 So I think I figured it out. Basically, I needed to note that $$C = D^{-1} s_{ij}$$ where$D^{-1}$is a diagonal matrix whose entries are$S_i$. Then, $$\lambda x = D^{-1} s_{ij} x$$ $$\lambda (D x) = s_{ij} x$$ $$\lambda (D x) = (s_{ij} D^{-1}) (D x)$$ Thus, the eigenvalues of$c_{ij}$are the eigenvalues of$s_{ij} D^{-1}$With$s_{ij} ...

0

Suppose that $u' D u$ is positive for every nonzero $u$. Consider a nonzero vector $y$, and look at $y' R y$. You can rewrite this as $(y'T') D (Ty)$. And now, letting $u$ denote $Ty$, you have $u' D u$, which you know to be positive. (Why? Because the matrix $T$ is invertible, so $Ty$ cannot be the zero vector, so the first sentence tells you it's ...

1

In general it is not sufficient to check the characteristic polynomial to make sure that two matrices are similar. In order to be similar, there needs to exist an invertible matrix $P$ such that $A = P^{-1}B P$. If two matrices are similar and one of them is diagonalizible (say, $B=Q^{-1}DQ$), then $A$ is automatically diagonalizible, too (by means of ...

1

$\lambda_1=-\varepsilon+t\sqrt{f_1f_2}$, hence $A-\lambda_1I=\begin{pmatrix}-t\sqrt{f_1f_2}&tf_1\\tf_2&-t\sqrt{f_1f_2}\end{pmatrix}$. Now, let $v=\begin{pmatrix}v_1\\v_2\end{pmatrix}$, so$$\left(A-\lambda_1 I\right)v=\begin{pmatrix}-t\sqrt{f_1f_2}v_1+tf_1v_2\\tf_2v_1-t\sqrt{f_1f_2}v_2\end{pmatrix}=\vec{0}$$ Now we can solve the equations and get ...

1

A graph with least eigenvalue at least $-1$ is a disjoint union of cliques. (Proof: the least eigenvalue of $K_{1,2}$ is $-\sqrt2$, interlacing.) The graphs with least eigenvalue at least $-2$ were characterized by Cameron, Goethals and Seidel. They are line graphs, so-called generalized line graphs, and a finite set of graphs associated to $E_6$, $E_7$, ...

1

If $\chi_A(A)$ is $0$ on each vector of a basis, $\chi_A(A)$ is $0$ on the whole space.

0

If we write $A=PDP^{-1}$ then it is not hard to show that $q(A)=Pq(D)P^{-1}$ for any polynomial $q$, so the problem reduces to the case where $A$ is itself diagonal. Can you see why the result is true for diagonal matrices?

0

I do not know much about eigenvalues or characteristic polynomials of matrices, but the identity permutation is the permutation that keeps the order of everything the same. It's called the identity because that's the naming convention things that keep everything the same. For example, $0$ is the identity of addition and $1$ is the identity of ...

3

Hint Consider the eigenvalues of $A \in SO(3, \Bbb R)$. In particular, (1) their product is $\det A = 1$, (2) they all have modulus $1$, and (3) any nonreal eigenvalues come in complex conjugate pairs. Now, if $A \neq I$, what can you say about the $1$-eigenspace of $A$?

1

Try to do what the link you posted says: $$1=x^2+6xy+y^2=(x\;y)\begin{pmatrix}1&3\\3&1\end{pmatrix}\binom xy=\binom xy^tA\binom xy$$ Diagonalize orthogonally the matrix $\;A\;$ (it's possible because it is symmetric): $$\begin{vmatrix}x-1&-3\\-3&x-1\end{vmatrix}=x^2-2x-8=(x-4)(x+2)$$ Now eigenvectors: ...

4

User John Brevik gives as a hint: What is the characteristic polynomial of $$\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$$ How could you have come up with this hint yourself? Well, as you say, the characteristic polynomial only cares about eigenvalues, so you need to find two matrices which are not similar but which have the same eigenvalues. ...

2

Answer to your last question: because there are nontrivial Jordan canonical forms.

4

\begin{bmatrix} 1&1 \\ 0& 1 \end{bmatrix} \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} consider these two matrices.(Jordan Canonical form is the answer).

0

If the deviation is significant, then it likely you have made a simple programming error. Check to see that you can recover the original matrix using the computed eigenvalue decomposition. I would expect to see small errors in individual component, but nothing major. If your inner products are tiny, but not zero, then is is most likely the inevitable ...

1

The quick answer to your question: note that the only diagonalizable matrix whose eigenvalues are all $0$ is the zero-matrix, and that a rank $1$ matrix can have at most one non-zero eigenvalue. Another approach: Note that any rank $1$ matrix can be written in the form $uv^T$ for column vectors $u,v$, and that a rank-$1$ matrix will be symmetric if and ...

3

The minimal polynomial of $A$ divides $x^4 - 4x^2 = x^2(x-2)(x+2)$ and the characteristic polynomial and the minimal polynomial have the same irreducible factors (possibly with different multiplicities), so a priori, you can only tell that the possible eigenvalues are $0,\pm 2$ and not necessarily all must occur (for example, the matrix $A = cI$ where $c \in ... 0 The matrix is diagonalizable for the spectral's theorem. Indeed the dimension of the eigenspace for the eigenvalue of$0$is$n-1$. Note that the eigenspace for the eigenvalue of$0$is the$\ker$of function. 1 Normal means precisely that$AA^\ast =A^\ast A$. Then $$\langle Av,Av\rangle=\langle v,A^\ast Av\rangle=\langle v,AA^\ast v\rangle=\overline{\langle AA^\ast v,v\rangle}=\overline{\langle A^\ast v,A^\ast v\rangle}$$ This gives the first result. Can you deduce the second from the first? 1 To find the generalised eigenvector you simply have to solve for $$\begin{bmatrix}\frac{1}{2} & -\frac{1}{2} \\ \frac{1}{2} & -\frac{1}{2}\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix} = \begin{bmatrix}1 \\ 1\end{bmatrix}$$ If$A$is your given matrix, this indeed means a solution$e_2satisfies $$(A+I)e_2=e_1,\quad\text{whence}\quad A ... 2 Your matrix, as it is written, is singular: it must be that zero is one of its eigenvalues. In fact, only zero is an eigenvalue of algebraic order two, and the homogeneous system to obtain its eigenspace is$$-\frac12x+\frac12y=0\implies x=y\implies V_{\lambda=0}=Span\left\{\binom11\right\}$$and it's dimension is one, thus there is no other eigenvector ... 1 In general, if p(x) is a polynomial an A a diagonalisable matrix, then so is p(A). First, note that if A=U^{-1}DU, then A^n=U^{-1}D^nU, and hence$$ p(A)=\sum_{k=0}^n c_kA^k=\sum_{k=0}^n c_kU^{-1}D^kU= U^{-1}\left(\sum_{k=0}^n c_k D^k\right)U. Clearly, \sum_{k=0}^n c_k D^k is also a diagonal matrix. 4 For the record, a counterexample to 1 (which you correctly disproved) is A=2I: A^4=16I=8A. If A=P^{-1}DP, substitute that into A^2-4A+8I, you get: \begin{align*} A^2-4A+8I={}&(P^{-1}DP)^2-4P^{-1}DP+8I=(P^{-1}DP)(P^{-1}DP)-P^{-1}\cdot4D\cdot P+8I={} \\ {}={}&P^{-1}D(PP^{-1})DP-P^{-1}4DP+P^{-1}8IP=P^{-1}(D^2P-4DP+8P)={} \\ ... 0 The key is to realize that you multiply partitioned matrices together in exactly the same manner by you do "regular" matrix matrix multiplication. The only difference is that multiplication is no longer commutative because your are dealing with submatrices rather than scalars. I shall demonstrate. Let A and B be partitioned conformally, i.e. ... 1 Recall that the set of all eigenvectors of matrix A corresponding to eigenvalue \lambda=2 V_2=\{ [\frac{k_2}{2} - \frac{k_1}{2}, k_2, k_1] \mid k_1,k_2 \in \mathbb R\}$$forms a vector subspace of \mathbb R^3. Since every element of V_2 can be expressed as a linear combination$$ [\frac{k_2}{2} - \frac{k_1}{2}, k_2, k_1] = k_1 ... 1 The formula $$X=k_1[-1/2,0,1]+k_2[1/2,1,0]$$ tells you, that for each cobination ofk_1,k_2$, the resulting vector satisfies$Ax=\lambda_1x$where$\lambda_1=2$. Therefore the eigenvectors for$\lambda_1$are$[-1/2,0,1]$and$[1/2,1,0]$The previous step $$[k_2/2-k_1/2,k_2,k_1]= k_1[-1/2,0,1]+k_2[1/2,1,0]$$ is basically a factorization. ... 2 The matrix$A$is real and symmetric, so it must be diagonalisable. Therefore the minimal polynomial (supposedly called$m_A$) has simple roots. The matrix$A+2I$has rank$1$and trace$6$, so its characteristic polynomial is$X^2(X-6)$, and the characteristic polynomial of$A$(supposedly called$f_A$) is obtained from it by substituting$X+2$for$X$: ... 1 You could just expand everything out in series, for example the first term goes like $$( \bar A ^T \bar A ) = ( A^TA+E_1^T A +A^TE_1 + E_1^TE_1)$$ $$( \bar B ^T \bar B ) ^{-1} = ( B^TB+E_2^T B +B^TE_2 + E_2^TE_2)^{-1}=(B^TB)^{-1}( I+E_2^T B(B^TB)^{-1} +B^TE_2(B^TB)^{-1} +E_2^TE_2(B^TB)^{-1} )^{-1}$$ Then use $$(I + C)^{-1} \approx I - C$$ you may use ... 0 Linear transformations do not change the dimension of the vector space. They just re-orientate & re-size the vector within the same space. A vector p = p(x) in this space has a max power (degree) of n. So its linear transform - here q = q(x) - must have no higher order than n also, though its coefficients must in general change. (I'm not sure why the ... 0 Hint: You can represent your transformation$T$as a matrix$A$w.r.t. to some choice of a basis of polynomials. Find the eigenvectors of$A$and translate them back to polynomials. From$Ax = \lambda x$for some$\lambda \in \mathbb C$and a nonzero coordinate vector$x$of a nozero polynomial$p$, you can deduce that the coordinate vector of$Tp$will be ... 0 No, It's not true in general. Let$A=S_1\Lambda_1S_1^{-1}$and and$B=S_2\Lambda_2 S_2^{-1}$, Then$AB=S_1\Lambda_1S_1^{-1}S_2\Lambda_2 S_2^{-1}$. So if$S_1=S_2$then$AB=S_1\Lambda_1S_1^{-1}S_1\Lambda_2 S_1^{-1}$$=S_1\Lambda_1\Lambda_2 S_1^{-1} and we know that product of two diagonal matrix is a diagonal matrix. Next the question is when, S_1=S_2 In ... 1 Recall that \lambda is an eigenvalue of A if \det(A-\lambda I) = 0. This means A - \lambda I is singular and therefore$$(A - \lambda I)v = 0$$has a non-trivial solution v \neq 0 called an eigenvector of A with respect to \lambda. With this, for any t \in \mathbb{F}, we have$$(A-\lambda I) (tv) = t (A-\lambda I) v = 0$$which shows tv ... 0 An eigenspace is a linear subspace, so there is never just one eigenvector: you can always at least takes its nonzero multiples as well (sometimes the dimension is more than~1, and your eigenvectors are arbitrary combinations of more than one chosen eigenvector). The linear system you solve to find eigenvectors is homogeneous, so that its solutions form a ... 0 An eigenvector \vec r of a matrix A is defined as vector which fulfils the equation$$ A\vec r = \lambda \vec r$$with the eigenvalue \lambda. So only the direction of the vector matters. Not its lenght. Thus \vec r'=k\vec r is also an eigenvalue of A because.$$ A\vec r' = k\cdot A\vec r = k\cdot \lambda \vec r=\lambda\vec r' So you can use ...

4

That's not true in general. let $A=\begin{pmatrix} 1&1\\0&\frac12 \end{pmatrix}$, $B=\begin{pmatrix}1&1\\0&2 \end{pmatrix}$, those 2 matrices are clearly diagonalizable since they have distinct eigenvalues, while $AB=\begin{pmatrix}1&3\\0&1\end{pmatrix}$ isn't.

1

You should solve $\left[\begin{array}{cc}0.5 & 0 \\ 2 & 0.5\end{array}\right]\left[\begin{array}{c}x \\ y\end{array}\right] = 0.5\left[\begin{array}{c}x \\ y\end{array}\right]$ That gives $0.5 x = 0.5x$ $2x + 0.5y = 0.5y$ So the only restriction is $x = 0$. You have only 1 eigenvector, and you are not getting 2 because the matrix is ...

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