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10

Since every polynomial has a root over $\mathbb{C}$, the characteristic polynomial of any complex matrix must have a root, say $\lambda$. Then $\lambda$ is an eigenvalue of the matrix at hand, which means the matrix has a non-zero eigenvector. Note that invertibility has nothing to do with the previous reasoning.


4

No, it does not make sense to talk about a Jordan canonical form for a linear map $\varphi:V\to W$. Alternatively, one could say it only makes sense once you choose a particular isomorphism $\psi:W\xrightarrow{\;\cong\;}V$, at which point what you're really talking about is the Jordan canonical form of the linear map $(\psi\circ \varphi):V\to V$. (In ...


3

The answer is no in general since Hermitian matrices have real eigenvalues. So for example, there is no Hermitian matrix whose characteristic polynomial is $X^2+1$. Even if you restrict to polynomials with real roots, I doubt you can find a simple formula : I think that if there is a rational formula that works for every polynomial with real roots, it ...


3

This is not true. The matrix $$ B=\pmatrix{ -2& 1\\ -22& 10} $$ has real and positive eigenvalue (matlab says $0.2583$ and $7.7417$), but $$ B+D=\pmatrix{ -1& 1\\ -22& 30} $$ has a negative eigenvalue (eigenvalues are $-0.2733$ and $29.2733$). The claim is true if in addition $B$ is symmetric, as Marcus M's answer shows.


3

Since $\mathbb{C}$ is algebraically closed, any matrix $A$ can be written (by change of basis) in Jordan normal form. Each Jordan block has exactly one eigenvector associated with it, so there is at least one eigenvector. (Worst case scenario, there is only one block; for example, this is the case with the matrices $\begin{bmatrix} 0 & 1 & 0 \\ 0 ...


3

If two matrices commute and are diagonalizable, then they can be simultaneously diagonalized by a common basis of eigenvectors. In this case, the eigenvalues of the product are the products of the eigenvalues of the two matrices for each common eigenvector. I think that beyond that, indeed this is a very difficult question, even if you assume one matrix is ...


3

The block matrix (let us denote by $M$) can be expressed as the Kronecker product of matrices $A$ and $I$ (the fixed size identity matrix, of dimension $n$) as follows:- $$M=A\otimes I$$ where $A$ is the $3\times3$ matrix:- $$A=\left[\begin{array}{ccc} \frac{3}{4} & \frac{1}{4} & \frac{1}{4} \\ \frac{1}{4} & \frac{3}{4} & -\frac{1}{4} \\ ...


2

The first question only requires the definition of an eigenvalue. For the second one, since $g\in \mathbb{C}[X], g(X)=c\prod\limits_{i=1}^n(X-\lambda_i)$ so $g(A)=c\prod\limits_{i=1}^n(A-\lambda_i I)$. And since the product of invertible matrices is invertible, you can conclude using the first point.


2

Note that in the case of $\lambda = 0$, we have $\det(A - 0\cdot I) = 0$; this implies that $0$ is an eigenvalue of $A$, i.e. not all of the eigenvalues are positive. However, it is the case that all eigenvalues of $A$ are non-negative. Since $B$ and $D$ are positive definite, then so is their sum, i.e. their sum only has positive eigenvalues. Now, if ...


2

We wish to compute $\det(\lambda I-A)$ where \begin{align*} I&=\begin{bmatrix}1&0\\0&1\end{bmatrix} & A&=\begin{bmatrix}3&0\\8&-1\end{bmatrix} \end{align*} Keeping the formula $$ \det\begin{bmatrix}a&b\\ c&d\end{bmatrix}=ad-bc $$ in mind, we may compute our determinant directly \begin{align*} \det(\lambda I-A) ...


2

Let $A\in M_n(\mathbb{R})$. $A$ has $n$ distinct real eigenvalues iff there are $\lambda\in\mathbb{R}$ and $P,Q$ real symmetric $>0$ s.t. $(A+\lambda I)^2=PQ$ AND the dimension of the commutant of $A$ is $n$.


2

If we have a $n\times n$ tridiagonal Toeplitz matrix of the form: $$A = \begin{bmatrix} a & c & & & & \\ b & a & c &&\mathbf 0 \\ & b & a & c \\ &&\ddots&\ddots&\ddots& \\ &\mathbf 0&&&& \\ &&&&&&&\end{bmatrix},$$ its eigenvalues are given ...


1

I just wanted to point out that linear transformations between different spaces do have a very nice form: If $T:V\to W$ then we can let $w_1, \dots, w_r$ be a basis for the range of $T$, and we can extend with vectors $w_{r+1}, \dots , w_m$ to a basis for $W$. Then we can let $v_1, \dots, v_r$ be such that $Tv_i=w_i$. Note that this implies that $v_1, \dots ...


1

For the first part, just note that $v^Tu \in \Bbb{R}$. Thus we have $$ Au = (I + uv^T)\ u = u + (uv^T)u = u + u(v^Tu) = (1 + v^Tu)\ u $$ which by definition means that $u$ is an eigenvector, corresponding to the eigenvalue $1 + v^Tu$. For the second part, first extend $v$ to a basis $\{v,u_1,\dotsc,u_{n-1}\}$. Applying the Gram-Schmidt process we can make ...


1

Notice, method is straight forward $$\lambda I=\lambda\begin{bmatrix}1&0 \\0&1 \end{bmatrix}=\begin{bmatrix}\lambda&0 \\0&\lambda \end{bmatrix}$$ & $$A=\begin{bmatrix}3&0 \\8&-1 \end{bmatrix}$$ $$\implies \lambda I-A=\begin{bmatrix}\lambda&0 \\0&\lambda \end{bmatrix}-\begin{bmatrix}3&0 \\8&-1 ...


1

Yes, eigenvectors corresponding to different eigenvalues are necessarily independent so, in an inner-product space, orthogonal. If an n by n matrix is symmetric then there are $n$ independent eigenvectors. Therefore, there exist eigenvectors, each of which spans a one dimensional vector space so the eigenvectors form an orthogonal basis for the entire ...


1

Hint: The null space of $A$ is always orthogonal to the range of $A^T$, for any matrix $A$. Now, exploit that $A$ is symmetric. Solution: Let $x$ be in the null space of $A$ and $y=A^T u$ be in the range of $A^T$. Then, we have $$ x^T y = x^T A^T u = (Ax)^T u = 0^T u = 0. $$ Since $x$ and $y$ were arbitrary, it follows that the null space of $A$ and the ...


1

Your intuition is wrong; singular values are "nice" that way. In particular: suppose that $A$ can be divided as $$ A = \pmatrix{A_0 & B\\C & D} $$ The interlacing property compares the singular values of $A$ to the singular values of $A_0$. That is, we are comparing the eigenvalues of $A^*A$ to the eigenvalues of $A_0^*A_0$. However, $A^*A$ has ...


1

An eigenspace has dimension greater than zero by definition. "Zero is always zero": Well yes, zero is always zero. So? When $\lambda=0$ in your example the dimension of the eigenspace is $1$ or $2$; that doesn't say zero is not zero, because the eigenvalue is not the dimension of the eigenspace. Why is the definition that way? The definition of eigenvalue ...


1

We start from $$A^N = \lambda_1 I + \lambda_2 A$$ Suppose $\mathbf{v}$ is the eigen vector corresponding to the eigenvalue $q_1$, then $$A\mathbf{v}=q_1\mathbf{v}.$$ From the first equation it follows: $$A^N \mathbf{v} = \lambda_1 I\mathbf{v} + \lambda_2 A\mathbf{v}.$$ Using the second equation, we get $$(q_1^N - \lambda_1 - \lambda_2q_1) ...


1

$A$ and $A^T$ are necessarily similar. To show this, it suffices to note that $$ \dim \ker [(A - \lambda I)^k] = \dim \ker [(A^T - \lambda I)^k] $$ for all $\lambda$ taken from the algebraic closure of $K$ and all $k \in \Bbb N$.


1

One way is to use eigen decomposition. If A is diagonalizable matrix then $ A = Q^{-1} \Lambda Q $ where $\Lambda$ is diagonal matrix with eigevalues on the diagonal. Then $ A^k = Q^{-1} \Lambda^k Q $ and $g(A) = Q^{-1} g(\Lambda) Q $. $g(\Lambda)$ is diagonal matrix with $g(\lambda_i)$ elements. From this place your proposition is easy to prove. ...


1

There is no precise answer to your question. Indeed, let $A,B\in M_n(\mathbb{C}),spectrum(A)=(a_i),spectrum(B)=(b_i),spectrum(A+B)=(c_i)$. Then the possible values of $(a_i,b_i,c_i)$ are dense in the set $\sum_i a_i+\sum_i b_i=\sum_i c_i$. Since we reason by density, we may assume that $B$ is the diagonal $D=diag((b_i))$. If $A,B$ are real, then we must add ...


1

The first requirement if such a generalization is to be acquired is to have an analogous notion of indices. However, for matrix with real entries, $A$ is symmetric if and only if $A$ is congruent to a diagonal matrix whose diagonal entries are $0$, $1$, and $-1$. If $A$ is not symmetric, we cannot conjugate it to get the indices. Another stumbling block of ...


1

If 0 is an eigenvalue of one of the matrices, it will also be an eigenvalue of the product, regardless of commutation. This is simple to see. Suppose A has an eigenvalue of 0. Then $\det A=0$. Since $\det AB=\det A \det B$, $\det AB =0$. So AB also has an eigenvalue equal to zero.


1

Counterexamples: - with $A=I_2$ the identity matrix and $\Sigma^{-\frac{1}{2}}=\operatorname{diag}\left(\frac{1}{2},2\right)$ we get $A^{\top}A=I_2$ and $B^{\top}B=\operatorname{diag}\left(\frac{1}{4},4\right)$. Thus, both matrices have different eigenvalues. - with $A=\left(\begin{smallmatrix}1&1\\0&1\end{smallmatrix}\right)$ also all eigenvectors ...


1

Of course $\lambda\not=\lambda_1$. By the power method, you can obtain the greatest eigenvalue $\lambda_2=tr(B^TB)-\lambda_1 $ of $B^TB$ and an associated eigenvector $v_2$. Let $P=[v_1,v_2]\in M_2$; then you know $B^TB=Pdiag(\lambda_1,\lambda_2)P^{-1}$; finally you must remove the scaling: $A^TA=\Sigma^{1/2}Pdiag(\lambda_1,\lambda_2)P^{-1}\Sigma^{1/2}$.



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