# Tag Info

6

No; consider the matrices $A=B=\begin{pmatrix}2&0\\0&2\end{pmatrix}$, or more generally $A=B=\lambda I$ with $\lambda\neq0,1$.

4

Hint: a rotation that is neither $180^\circ$ nor $0^\circ$ has no real eigenvalues.

3

You have to reconstruct the (right) eigenbasis of $A$ first. Since two eigenvectors are given, there is enough information to find the second right eigenvector. Let $v$ denote this eigenvector, then $Ap_{r2}=-p_{r2}$. Multiplying this from the left by $p_{l1}^T$, we get $$-p_{l1}^Tp_{r2} = p_{l1}^TAp_{r2} = 2p_{l1}^Tp_{r2},$$ which proves ...

3

Suppose $T$ is a bounded operator on a Banach space $X$. $\lambda\in\rho(T)$ iff $T-\lambda I$ is a linear bijection. In that case, the inverse $(T-\lambda I)^{-1}$ is automatically continuous by the closed graph theorem. There are three basic things that can stand in the way of $T-\lambda I$ being invertible. $T-\lambda I$ is not injective. Equivalently, ...

3

The typical approach to this problem is not to show directly that an eigenvalue with multiplicity $m$ for a symmetric matrix has an $m$-dimensional space of corresponding eigenvectors but to use an inductive argument which shows it indirectly. In order to do that, it is more comfortable to talk about self-adjoint maps instead of real symmetric matrices ...

3

First note that there is an equivalent definition for characteristic value - a characteristic value of $A$ is a scalar $c$ such that the matrix $(A-cI)$ is NOT invertible. For (i) - if $0$ is a characteristic value of $AB$ then $AB$ is not invertible $\Rightarrow BA$ is not invertible and hence $0$ is a characteristic value of $BA$. Let us assume that ...

3

In the general case, $\lambda_2$ and $\lambda_3$ are both nonzero. When $\lambda$ is one of them you know that $$\left|\begin{matrix}-\lambda & 1 & 0 \\ -f'(u_1) & -c-\lambda & 1 \\ 0 & 0 & -\lambda \end{matrix}\right| = -\lambda \left|\begin{matrix}-\lambda & 1 \\ -f'(u_1) & -c-\lambda \end{matrix} \right| = 0$$ so the ...

3

A question you might wish to answer first: What is the relationship between the eigenvalues of $A$ and the eigenvalues of $A^3$?

2

Well, one way to look at it is considering the identity $$e^{i\theta}=\cos(\theta)+i\sin(\theta)$$ which tells us that the function $e^{i\theta}$ traces out the unit circle of the complex plane. One might note that if we have an eigenvalue of the form $e^{i\theta}$ then powers of it are of the form $e^{ni\theta}$ - so they are just rotating around the unit ...

2

Every square matrix is similar to a matrix in what's called Jordan Canonical Form. This has various properties, but most important here is that it is upper triangular, and the eigenvalues (of both the new and original matrix) are on the diagonal of the resulting matrix. The way to think about this process is that we change bases, and in that new basis ...

2

$P_{\lambda}(A^{T})=\det(\lambda I_{n}-A^{T})= \det((\lambda I_{n}-A)^{T})= P_{\lambda}(A)$

2

I believe this part went wrong: $$-Q-\lambda+Q\lambda+\lambda^2+P^2 = \lambda^2+(Q-1) \lambda + P^2 - Q \ne \lambda^2+(1-Q)\lambda+P^2 - Q$$

2

You are assuming that $P$ is invertable but it need not be. However your conclusion about the eigenvalues is correct. If $\lambda$ is an eigenvalue of $P$ with corresponding eigenvector $v$ then $PPv = P\lambda v = \lambda^2v$ but also $PPv = Pv = \lambda v$. Thus $\lambda^2 = \lambda$ from which follows $\lambda=0$ or $\lambda=1$. In general matrices $P$ ...

2

Your lecturer is correct: there is a lot wrong with the proof above. The result (that every eigenvalue of a real symmetric matrix is real) has nothing to do with the Fundamental Theorem of Algebra---if the characteristic polynomial had no complex roots, then there would be no eigenvalues and the claimed result would be trivially true. The quantifier in the ...

2

Consider $-I$, $(-I)^2=I$ ( we suppose $-I=A=B$) and -1 is an eigenvalue of -I but not of $I$

2

Hint: Calculate $y'Ax$ two different ways, and relate the answer to $y'x$.

1

In control theory and dynamical systems you have modal decomposition, which is a very useful tool to quickly create the dynamic equation for a given (real life) system Given a system of differential equation: $\dot x(t) = Ax(t)$, $x(0) = x_o$, $A$ has distinct eigenvalues Then the solution to this equation is given as: $x(t) = \sum\limits_{i=1}^n ... 1 For$T$to have$\lambda$as an eigenvalue,$T-\lambda I$must be non-injective. For$\lambda$to be in the spectrum of$T$, it must only be non-invertible. These are equivalent when$T$is an operator on a finite-dimensional space, but not in general! For example, let$T$be the shift operator$(x_0,x_1,\dots) \mapsto (0,x_0,x_1,\dots)$on your favorite ... 1 This is fairly standard stuff. If$A v_k = \lambda_k v_k$then let$V = \begin{bmatrix} v_1 & \cdots & v_n\end{bmatrix}$and$\Lambda = \operatorname{diag}(\lambda_1, \cdots , \lambda_n )$. Note that$V$is invertible. Then the above equations can be written as$AV = V \Lambda$which gives$V^{-1} A V = \Lambda$. If$A$is similar to a diagonal ... 1 Since$A$is positive semi-definite,$A^{1/2}$exists and is Hermitian too. Since$tr(AB)=tr(BA)$$$tr(AB)=tr(A^{1/2}A^{1/2}B)=tr(A^{1/2}BA^{1/2})$$ Since $$(A^{1/2}BA^{1/2})^*=(A^{1/2})^*B^*(A^{1/2})^*=A^{1/2}BA^{1/2}$$ it is Hermitian and thus all its eigenvalues are real. So$tr(A^{1/2}BA^{1/2})$is real. 1 This seems to address your problem, I found a very nice basis for the eigenvectors of a matrix with all entries$1.$The reason we know the columns are independent is that they are perpendicular to each other, ordinary dot product of columns is zero. I am encouraging you to do something along these lines. $$\left( \begin{array}{rrrrrrrrrr} 1 & ... 1 Suppose A(x)=cx, B(A(x))=cB(x)=A(B(x)), this implies that B(x) is an eigeinvector of A associated to c, since A has n distinct eigenvalues, the eigenspace associated to c has dimension 1, thus B(x)=dx. 1 Hint: the determinant is equal to: \operatorname{det}(A-\lambda I) = (1-\lambda)\operatorname{det}\begin{pmatrix} 2-\lambda & c\\ c & 3-\lambda \end{pmatrix} + (-1) \cdot a \cdot \operatorname{det} \begin{pmatrix} a & c \\ b & 3- \lambda \end{pmatrix} + \dots Can you finish it yourself? 1 This inequality does not hold if M is a positive scalar multiple of the identity matrix. More generally, this inequality does not hold (strictly) if all the column sums of M are equal, i.e., if \mathbf{1}^TM = \alpha\mathbf{1}^T, for some positive scalar \alpha < s. To see this, observe that$$ \mathbf{1}^T(sI - M) = (s - \alpha)\mathbf{1}^T ... 1 What can be said about eigenvalues of$A+E_1$? Not very much, I think. Just consider$A=\begin{bmatrix}0&1\\1&0\end{bmatrix}$. This matrix leaves alone the span of$\begin{bmatrix}1\\1\end{bmatrix}$, and negates the span of$\begin{bmatrix}-1\\1\end{bmatrix}$. So its eigenvalues are$1$and$-1$. But ... 1 The matrix$A+E_{1,1}$is still real symmetric. It can in fact be any real symmetric matrix that has at least one nonzero off-diagonal entry in its first column (because of the non-commutation condition). What can be said about its eigenvalues and eigenvectors is precisely what can be said about them for an arbitrary real symmetric matrix (real eigenvalues, ... 1 Suppose that$\lambda \neq 1$is an eigenvalue. Then $$\pmatrix{1 - \lambda & -P\\ P & -Q - \lambda}$$ is singular, so that the top and bottom rows are multiples. Verify that the vector $$v = \pmatrix{P \\ 1 - \lambda}$$ must be an eigenvector associated with$\lambda$. 1 If$A$is similar to$B$then there exists invertible$P$such that$A=P^{-1}BP$. We can rewrite this as$PA=BP$. Suppose$\lambda$is an eigenvalue of$A$with eignevector$v$. Then $$PAv=P(\lambda v)=\lambda Pv=BPv$$$\therefore\lambda$is an eigenvalue of$B$with eigenvector$Pv$. Switching the roles of$A$and$B$gives you that an eigenvalue of ... 1 yet another one. Since$A$and$B$are similar, there exists$P$such that$B=PAP^{-1}$. $$\det(A-\lambda I)=\det(P)\det(A-\lambda I)\det(P^{-1})= \det(B-\lambda I)$$ Thus$A$and$B\$ share same eigenvalues and characteristic polynomial.