# Tag Info

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 ...

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 mapL$... 4 Your matrix is a strictly upper triangular matrix. Upper triangular matrices$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$. 4 The asserted equality of two determinants is incorrect. Let$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 ... 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 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$k\geq2$. 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 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 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 ... 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 ... 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 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 ... 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] = ...

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

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 ...

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 $A^3=I$.

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

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

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 ...

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 $A(C)$.

1

Notice, method is straight forward $$\lambda I=\lambda\begin{bmatrix}1&0 \\0&1 \end{bmatrix}=\begin{bmatrix}\lambda&0 \\0&\lambda \end{bmatrix}$$ & $$A=\begin{bmatrix}3&0 \\8&-1 \end{bmatrix}$$ $$\implies \lambda I-A=\begin{bmatrix}\lambda&0 \\0&\lambda \end{bmatrix}-\begin{bmatrix}3&0 \\8&-1 ... 1 I just wanted to point out that linear transformations between different spaces do have a very nice form: If T:V\to W then we can let w_1, \dots, w_r be a basis for the range of T, and we can extend with vectors w_{r+1}, \dots , w_m to a basis for W. Then we can let v_1, \dots, v_r be such that Tv_i=w_i. Note that this implies that v_1, \dots ... 1 Given$$ \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. $$From (OP) follows$$ ...

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

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

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 $k \in \Bbb N$.

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