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17

Hint. $$A^3-A+I=0\quad\Longrightarrow\quad A(I-A^2)=I.$$

7

Let $v\neq 0$ be an eigenvector of $T$ with eigenvalue $\lambda$, so $Tv=\lambda v$. Using $T=T^2$ we have $$Tv = T^2 v = T(Tv) = T(\lambda v) = \lambda(Tv) = \lambda^2 v.$$ Hence, $\lambda v = \lambda^2 v$. Since $v\neq 0$ we conclude $\lambda = \lambda^2$. The only solutions to this equation are $0$ and $1$.

6

Think of this as follows: $$T^2=T\implies T(T-I)=0$$ Thus, $\;T\;$ is a root of $\;x(x-1)\;$ and thus the characteristic polynomial of $\;T\;$ can only have $\;0\;$ or $\;1\;$ as its roots, and thus these precisely are the only possible eigenvalues of $\;T\;$ . So you were half right...:)

5

Knowing just the eigenvalues of a matrix is not enough to tell whether it is magic. For example, the real matrices $$\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \qquad \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$$ both have the eigenvalues $1$ and $-1$, but only the second of them is magic. So you need to adopt a more liberal ...

5

The space of polynomials is infinite-dimensional. So there is no characteristic polynomial, and no Cayley-Hamilton theorem. If $g$ is a polynomial of degree $n$, then $g(D)$ is the differentiation of order $n$. If you take a polynomial $p$ of degree $n+1$, then $g(D)p\ne0$, hence $g(D)\ne0$. On the other hand, if you consider the space of polynomials of ...

4

Call $B= (-1)(A^2 - I)$, now $(-1)(A^2 - I)A=I$ so you have $B$ as one side inverse, on the other side also $A(-1)(A^2-I) =I$.

4

Hint: For any square matrix of order $n$, the constant term of the characteristic polynomial $\chi_A$is $(-1)^n\det A$, and the minimal polynomial of $A$ divides $\chi_A$.

3

There exists a polynomial $p(t)$ such that $$\DeclareMathOperator{char}{char}\char_A(t)=p(t)m_A(t)$$ It follows that $$(-1)^n\det(A)=\char_A(0)=p(0)m_A(0)=p(0)b_0$$ That is, $$\det(A)=(-1)^np(0)b_0$$ Hence $b_0$ divides $\det(A)$.

3

Hint : The minimal polynomial of the matrix $A$ must be a divisor of the polynomial $$x^3-x^2-3x+2=(x-2)(x^2+x-1)$$

3

Hint If $\lambda$ is an eigenvalue of $A$, then $\lambda^2$ is an eigenvalue of $A^2$. (Also, if $A$ is real but you want to include complex eigenvalues, it's useful to know that if $\lambda$ is an eigenvalue of a real matrix $A$, then (1) so is its complex conjugate $\bar{\lambda}$, and (2) $\lambda$ and $\bar{\lambda}$ have the same multiplicity.)

3

Hint: $D^{-1}=D$ since $F=\mathbb{F}_3$ and $D$ is diagonal.

3

Let $A$ be $m\times m$ matrix such that sum of each row is equal to $n$, then consider $$\begin{pmatrix}a_{11}&a_{12}&\cdots &a_{1m}\cr a_{21}&a_{22}&\cdots &a_{2m} \cr \vdots & \vdots & \ddots & \vdots \cr a_{m1}&a_{m2}&\cdots &a_{mm}\end{pmatrix}\begin{pmatrix} 1\cr 1 \cr \vdots \cr 1 ... 3 I don't think the following could be caracterized as "nasty determinant calculation". I don't know how one can prove the equality without indulging in some computation. Let r=\operatorname{rank}(A) From a well-known theorem, derive that there exists P,Q invertible m\times m and n \times n matrices such that$$A=P\begin{bmatrix}I_r& 0\\ 0 ...

2

It is not equal to the largest absolute value of eivenvalues. For a counterexample, consider a matrix $A_{11}=0, A_{12}=A_{21}=A_{22}=1$. However, it is a bound of eigenvalues. Let $A$ be a complex square matrix. Then, all its eigenvalues are bounded by $min \{ max_j \sum_i |A_{ij}|, max_i \sum_j |A_{ij}|\}$. If you assume the matrix is symmetric, then ...

2

The system you propose is fine. You should find the following generalized eigenvectors: $$\vec{v}_3=\begin{bmatrix}0\\1\\0\\0\end{bmatrix},\quad\vec{v}_4=\begin{bmatrix}0\\0\\0\\1\end{bmatrix}$$

2

First out, definition of Generalized Eigenvector, So you would want to try and solve $({\bf A}-\lambda {\bf I})^k{\bf v} = {\bf 0}$, but $({\bf A}-\lambda {\bf I})^{k-1}{\bf v} \neq {\bf 0}$ to answer your question. You can use the first order eigenvectors (ordinary eigenvectors) together with the block-zero property to limit the span of generalized ...

2

For a route different to induction, express the matrix as $$A=aI_3+G$$ where $I_3$ is the identity matrix and $G$ is given by $$G=\begin{bmatrix}0&0&0\\1&0&0\\0&1&0\end{bmatrix}$$ A binomial expansion will yield \begin{align}A^k&=(aI_3+G)^k\\&={k \choose 0}a^kI_3+{k \choose 1}a^{k-1}I_3G+{k \choose ... 2 The system you derived does not force the vector to be the zero vector: it only forces the second component to be zero. The first component can be anything to satisfy the system, and anything nonzero to be an eigenvector. For convenience you can choose the first component to be 1. Incidentally, because the system is triangular, you don't need to use ... 2 Note that as 3 is odd, there must be at least one real eigenvalue. If there exists only 1 real eigenvalue, then that eigenvalue must be \pm 1 because then the inverse of the eigenvalue also has to be an eigenvalue of B (by supposition). Otherwise there must exist 3 real eigenvalues, call them x,y,z. Then WLOG suppose x=1/y as then x^{-1} and ... 2 Suppose v is an eigenvector of A, with eigenvalue \lambda. Then \lambda must satisfy the same cubic equation that A satisfies:(A^3-A^2-3A+2I)v=0v=0\\ (A^3-A^2-3A+2I)v=AAAv-AAv-3Av+2v\\ =AA\lambda v-A\lambda v-3\lambda v+2v\\ =\lambda A(Av)-\lambda(Av)-3\lambda v+2v\\ =\lambda A(\lambda v)-\lambda^2 v-3\lambda v+2v\\ ...

2

Another side remark: You say that you are not sure if 1, or both 0 and 1 can be eigenvalues. In some cases, it is worthwhile to think of specific examples and see what they can tell us. So what are some examples of matrices $T$ that satisfy $T^2 = T$? Well, the identity is certainly one, and its eigenvalues are all 1. However, another such matrix is the ...

1

Let $\lambda$ be an eigenvalue of $A$ with corresponding eigenvector $X$. We have $$AX=\lambda X.$$ Consider $$B=A-a I.$$ We have $$BX = (A-aI)X=AX-aIX=\lambda X-aX = (\lambda - a)X.$$ Thus, if $\lambda$ is an eigenvalue of $A$, then $\lambda-a$ is an eigenvalue of $B$.

1

If c is an eigenvalue of $A$, then $Av = c.v$ for the corresponding eigenvector. This implies $(A - aI)v = (c - a)v$. Thus, if c is an eigenvalue of $A$, then $c-a$ is an eigenvalue of $A - aI$. This also shows that eigenvectors of $A$ and $A - aI$ are same.

1

The eigenvalues of $A$ are the roots of its minimal polynomial, and $\bigl\{P(x)\in \mathbf R[x] \mid P(A)=0\bigr\}$ is the set of multiples of the minimal polynomial of $A$. Hence if $P(A)=0$, the eigenvalues of $A$ are among the roots of $P(x)$.

1

Check this out: Since $A^3 - A+ I = 0, \tag{1}$ we have $A(I - A^2) = (I - A^2)A = I. \tag{2}$ (2) shows that $A^{-1} = I - A^2, \tag{3}$ so $A$ is invertible. It is not in general true that (1) implies the characteristic polynomial of $A$ is $t^3 - t + 1$; if $\text{size}(A) \ne 3$, for example, it cannot be the case, since the degree of the ...

1

For any ring $R$, and any polynomials $f(x), d(x), q(x) \in R[x]$, we have $f(x) = d(x) q(x) \Rightarrow f_0 = d_0 q_0, \tag{1}$ where $f_0$ is the constant term of $f(x)$ etc. This is easy to see by simply examining the basic rule for polynomial multiplication, with $d(x) = \sum_0^{\deg d} d_i x^i, \tag{2}$ $q(x) = \sum_0^{\deg q} q_i x^i, \tag{3}$ ...

1

A more direct proof would be: Assume $\lambda$ is a non-zero eigenvalue of $f\circ g$, i.e. $$(f\circ g)v=\lambda v$$ for some $0\neq v\in V$. Application of $g$ on both sides shows $(g\circ f) (g(v)) =\lambda g(v)$ which shows that $\lambda$ is also an eigenvalue of $g\circ f$ (with eigenvetor $g(v)$, which is not equal to $0$, otherwise $(f\circ ... 1 This works not just for square matrices but all matrices where the product makes sense. A reference is Godsil and Royle's book Algebraic graph theory, Lemma 8.2.4. Their ingenious proof is to note that$$\det\left(\begin{bmatrix}I & A \\ B & I\end{bmatrix} \begin{bmatrix}I & 0 \\ -B & I\end{bmatrix}\right) = \det\left(\begin{bmatrix}I & ... 1 Because$A^2=-I$, the minimal polynomial of$A$has to divide$f(x)=x^2+1=(x+i)(x-i)$. So all of the possible eigenvalues for$A$are$\{i\}$,$\{-i\}$, or$\{i,-i\}$. 1 Recall for linear PCA$N$is the number of data points in the set and comes in the eigen decomposition of the covariance matrix$C$, where$C = \frac{1}{N} \sum_{i=1}^N x_i x_i^\top$( we look for eigenvectors$v$such that$\lambda v = C v$). So we simply multiply through by the constant$N\$ to bring it through to the other side. It is similar ...

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