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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 ...


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.


0

By Brauers Theorem ( Matrix Analysis by Horn & Johnson ) subtracting by $\lambda {\bf I}$ will reduce all eigenvalues by $\lambda$ and if none equals $\lambda$ means each of them will become non-zero.


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. ...


0

Hint: $$g(x)\in \mathbb{C}[x]\implies \sigma\left(g(A)\right)=\{g(\lambda):\lambda\in \sigma(A)\}$$ where $\sigma(\cdot)$ denotes the spectrum, i.e. collection of eigenvalues of a matrix.


0

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 ...


0

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: $A^TA=\Sigma^{1/2}Pdiag(\lambda_1,\lambda_2)P^{-1}\Sigma^{1/2}$.


0

Burnside's theorem. Let $K$ be an algebraically closed field. The algebra generated by $A,B\in M_n(K)$ is whole $M_n(K)$ iff $A,B$ have no common invariant proper subspace. Then, your assertion is false, except if $n=2$. EDIT. (Answer to Mariano). If $n=2+p\geq 3$, then let $K^n=E_2\oplus F_p=E_2\oplus G_p$ where $F_p\cap G_p=\{0\}$. Let $A$ be $E_2$ and ...


0

Even though this question is very old, I thought I would give an answer to show a different, more algebraic approach. Since $A$ is a normal matrix it can be shown that: 1) $||A\overrightarrow{z}|| = ||A^*\overrightarrow{z}||$ 2) $ A - \lambda I $ is normal Given that $\lambda$ is an eigenvalue with some corresponding eigenvector $\overrightarrow{z}$, ...


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

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 [b]independent[/b] 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 ...


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.


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 ...


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 $k \in \Bbb N$.


0

A first step could be the following: A corollary to Weyl's eigenvalue inequalities (see, e.g., Horn & Johnson, second edition, Corollary 4.3.15), is that the largest eigenvalue of a sum of matrices is as least as large as the sum of the largest eigenvalue of one of the matrices and the smallest eigenvalue of the other matrix, i.e., $$ ...


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

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

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 ...


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) ...


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 ...


0

Aren't the eigenvalues of $\Gamma$ always $1\pm \frac{1}{\sqrt{2}}$? I'm confused as to why you're asking for asymptotics (and why you say this is only interesting for odd $n$). If we express $\rho$ and $\sigma$ using coordinates in the $\omega$ basis, they are $R=Q_R \Lambda Q_R^*$ and $S=Q_S \Lambda Q_S^*$ where $$\Lambda = \left[\begin{array}{cc} ...


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.


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 ...


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 ...


2

First for the eigenvalues: By the spectral theorem (and since $A$ is a positive (hence self-adjoint) operator), there is a measure space $(X, M, \mu)$ and some (measurable) function $g : X \to [0,\infty)$, such that $A$ is unitarily equivalent to the multiplication operator $$ M_g : L^2(\mu) \to L^2 (\mu), h \mapsto g\cdot h. $$ Thus, it suffices to prove ...


1

Hint The argument so far does not use that $\lambda$ is an eigenvalue: Since $\lambda$ is an eigenvalue, it is a root of the characteristic polynomial.


1

$\lambda$ is a root of the characteristic polynomial $x^2 + ax + b$, hence $-a\lambda-b=\lambda^2$, and you are done.


4

The asserted equality of two determinants is incorrect. Let $M$ be the matrix whose determinant appears first; let $N$ be the matrix whose determinant appears after the "equals" sign. You multiplied the second row by $2-X$. That alters the determinant. Then you added $-1$ times the first row to the second, and that does not alter the determinant. Then ...


1

The trace of $M^n$ is $p_n = \sum_{\xi_i\in E}\xi_i^n$ where $E$ is the set of eigenvalues. The power sums are a base of the ring of the symmetric functions as well as the elementary symmetric functions. As a matter of fact, we may switch from one base to the other through Newton's identities. But the elementary symmetric functions of the eigenvalues give ...


2

Two diagonalizable matrices $A$ and $B$ will commute if and only if they are simultaneously diagonalizable; i.e., we can write $$A = P^{-1}D_BP$$ and $$B=P^{-1}D_BP$$ where $D_A$ and $D_B$ are diagonal, and the elements of $D_A$ and $D_B$ represent the eigenvalues of $A$ and $B$, respectively. Pointed out by Ian in the comments, a nondiagonalizable matrix ...


0

Could you be more precise about what you want to compute? The solutions $(\alpha,\beta,\gamma)$ and $t$ to the first equation do not form a countable set, since you can for instance fix (arbitrary) $\beta$ and $\gamma$ and get $2n$ solution pairs $(\alpha,t)$.


0

This problem is a subset of the problem called a multiparameter eigenvalue problem. The solution can be explicitly expressed as the solution to a standard (or generalized) eigenvalue problem using Kronecker products. Considerable theory was worked out by Atkinson in the 60's, but more recently popularized in numerical methods by for instance Plestenjak and ...


1

Given $$ \mathbf{A}_n^3 = \mathbf{I}_n \hspace{1em} \textrm{and} \hspace{1em} \mathbf{A}_n \ne \mathbf{I}_n. \tag {OP} $$ Eigenvalues Let $\lambda$ be an eigenvalue of $\mathbf{A}_n$. Therefore $$ \mathbf{A}_n \vec{x}_\lambda = \lambda \vec{x}_\lambda. $$ Whence $$ \mathbf{A}_n^k \vec{x}_\lambda = \lambda^k \vec{x}_\lambda. $$ From (OP) follows $$ ...


2

For clarity of concept, it is worth to note that, by writing $$Q = \left(\begin{array}{cccc} A&B\\0&C\end{array}\right)$$ $A,C$ are two square matrices, it implies that: 1.a. $Q$ is a linear transformation from one vector space $V$ into itself. 1.b. There $V_1$ and $V_2$ subspaces of $V$ such that $V=V_1\oplus V_2$. 1.c. $A$ is a linear ...


1

One way you could try this is to note $A^2=A^{-1}$. Then write $$A=\pmatrix{a&b\\c&d}$$ such that $ad+bc=1$. Then, $$A^2=\pmatrix{a^2+bc&b(a+d)\\c(a+d)&bc+d^2}$$ $$A^{-1}=\pmatrix{d&-b\\-c&a}$$ Then you see we need things like $a+d=-1$, etc. You could play with these and see if something comes out of it


1

You can even demand that $A$ have integer entries. For example, $$A=\begin{bmatrix} 0&1\\ -1&-1 \end{bmatrix}\,.$$


2

You just have to take the companion matrix of $\Phi_3(x)$, i.e. the third cyclotomic polinomial: $$ A = \begin{pmatrix}0&-1\\ 1&-1\end{pmatrix}.$$ We have $A,A^2\neq I$, but $A^3=I$.


3

Yes. There is a standard embedding homomorphism from the field of complex numbers to ring of real $2\times 2$ matrices: $$a+bi \mapsto \begin{pmatrix}a&b\\-b&a\end{pmatrix}$$ So find a complex cube root of $1$. There are actually cases with entries rational. For example: ...


3

Yes, consider $$ A=\begin{pmatrix} \cos2\pi/3 &\sin2\pi/3\\ -\sin2\pi/3& \cos2\pi/3 \end{pmatrix} $$ Minor modification gives an example for any other natural $k\geq2$.


8

Example $$A= \begin{bmatrix} \cos(120^\circ) & \sin(120^\circ) \\ -\sin(120^\circ) & \cos(120^\circ) \end{bmatrix}$$ Now, if $P$ is any invertible matrix, then $PAP^{-1}$ also has this property, with the above $A$. If you know about complex eigenvalues/eigenvectors, you can prove that any matrix with this property must have non-real eigenvalues ...


1

It is possible to show that $im$ $Q = L_1 + L_2$ and $Q|L_1 = A, Q|L_2 = C$ where $L_1$ and $L_2$ are invariant subspaces of $Q$. Since eigenspace of $\lambda$, $W_\lambda \subseteq L_1$ $(W_\lambda\subseteq L_2)$ it is obvious that $n = dim$ $W_\lambda$ is eigenspace dimension of $\lambda$ of both $Q$ and $A(C)$.


0

It is convenient to work with the standard basis $\left( \begin{matrix} 1 \\ 0 \end{matrix} \right), \left( \begin{matrix} 0 \\ 1 \end{matrix} \right)$ to write the answers directly as vectors in the standard basis. So we must solve, with $T= \left( \begin{matrix} a & b \\ c & d \end{matrix} \right)$ $$ \left( \begin{matrix} a & b \\ c & d ...


3

Left/right shift operators are the standard examples, but I personally think that the multiplication operator is the easiest way to see that there may be something else in the spectrum besides eigenvalues. Consider a multiplication operator $A_c$ on $\ell^\infty$ ($c\in\ell^\infty$) $$ (A_c x)_n=c_n x_n. $$ The inverse if exists is clearly a multiplication ...


7

If something is non-invertible, there's two (non-disjoint) possibilities: it fails to be injective, or it fails to be surjective. In finite dimension, these are the same, but in infinite-dimensional spaces, weird things can happen. If it fails to be injective, there's $x \ne y$ such that $(T - \lambda I)(x) = (T - \lambda I)(y)$. So $(T - \lambda I)(x - y) ...


6

For finite-dimensional vector spaces, injectivity and surjectivity are equivalent. That's not the case for an arbitrary Hilbert space. The classic examples are the left- and right-shift operators $L, R:\ell^2 \to \ell^2$, given by \begin{align*} L(x_1, x_2, \dots) &= (x_2, \dots) \\ R(x_1, x_2, \dots) &= (0, x_1, x_2, \dots). \end{align*} The map $L$ ...


3

$T-\lambda I$ being non-invertible does not imply there is a non-zero $x$ with $(T-\lambda I)x=0$. That is true when $H$ is finite-dimensional, but not necessarily when $H$ is infinite-dimensional. The classic counterexample is the right-shift operator $R:\ell^2(\mathbb{N})\to\ell^2(\mathbb{N})$. Take a look at the Wikipedia article on the notion of ...


2

This is too large for a comment, but I thought it might help to shed some light on the underlying reason that the matrix should be expected to be upper-triangular, and indeed nilpotent (so that you would sooner suspect an error in formatting, or perhaps in the choice of basis order). Your transformation (call it $L[f]$) can be broken down into $L[f] = ...


4

Your matrix is a strictly upper triangular matrix. Upper triangular matrices $n\times n$ are all nilpotent since their characteristic polynomials, $\det (A-XI_n)$, is equal to $(-1)^nX^n$. According to Cayley-Hamilton's Theorem, $(-1)^nA^n=0\Rightarrow A^n=0$.


3

Since $c_0=0$ $$f_T = \sum\limits_{i=0}^n c_ix^i=x\sum\limits_{i=0}^{n-1} c_{i+1}x^i=xg(x)$$ Since $c_1=1$, $x$ and $g(x)$ are coprime. The minimum polynomial $m_T$ must contain all factors that are divided by $f_T$ and so, $m_T=xh(x)$, where $h|g,\gcd(x,h)=1$. So by theorem of invariant factors for minimal polynomial, $V$ is the direct sum of invariant ...



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