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17

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


8

Another approach: The characteristic polynomial is $c(x) = x^3 - 2x^2 - 15x - k$. Using basic calculus, $p(x) = x^3 - 2x^2 - 15x$ has three distinct real roots with local max and min of $400/27$ and $-36$ (respectively at $x = -5/3$ and $x = 3$). Hence $c(x) = p(x) - k$ will likewise have three distinct real roots iff $k \in (-36,400/27)$.


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 ...


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 $A(-1)(A^2-I) =I$.


3

Hints: Note that the matrix $A$ has rank $2$. So, for $n>2$, $A$ will have eigenvalue $0$ with multiplicity $n-2$. Note that $A$ is symmetric. It follows that the kernel of $A$ is the orthogonal complement of its image. That is, the eigenspace associated with $\lambda = 0$ is the orthogonal complement of the space spanned by the vectors $$ ...


3

Hint: Find the characteristic polynomial. A polynomial has a multiple root if and only if its discriminant equals to $0$. Or, in other words, a polynomial has a multiple root if and only if it has a common root with its derivative.


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

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

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

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

There are two main cases for your question: 1) The description of the operator is specified in such a way that you would be unable to determine its action on any given basis. e.g. The mapping from $\mathbb{R}^2$ to itself that "is a rotation". There are many such rotations, and given the eigenvalues and eigenvectors, you know the transformation. So there ...


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

Unfortunately, the inequality you seek to prove seems not to be established for the general positive semi-definite matrices. I can only prove it for very specific case of $K_1, K_2$ such that eigenvalues of the sum of these matrices are always less than or equal to the corresponding sum of eigenvalues of each of them. Below is the proof for this ...


2

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 ...


2

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

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

There is no standard way of presenting a vector space other than by presenting a basis, and so there is no standard way of presenting a linear map between two vector spaces other than a matrix (or something that is equivalent to a matrix, e.g., defining the map on a large enough collection of vectors). So any situation where you are trying to compute ...


1

$\left( \begin{array}{ccc} 1-\lambda & -1 & 2 \\ -1 & 1-\lambda & 2 \\ 2 & 2 & 2-\lambda \\ \end{array} \right) $ = $\left( \begin{array}{ccc} 2-\lambda & -2+\lambda & 0 \\ -1 & 1-\lambda & 2 \\ 2 & 2 & 2-\lambda \\ \end{array} \right) =(-2+\lambda) \left( \begin{array}{ccc} -1 & 1 & 0 \\ -1 ...


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 ...


1

Since you don't have access to the cubic discriminant during the exam, it looks to me like consulting the characteristic polynomial and factorizing is your best bet. Ultimately this is an exercise in calculus. Luckily the matrix is given to you so that this polynomial isn't too bad. If my calculation is correct I find that $$\det(A-I\lambda) = ...


1

Suppose $A=\begin{pmatrix}b_1\\b_2\\\vdots\\b_n\end{pmatrix}$ and let $B =\begin{pmatrix}x_1\\x_2\\\vdots\\x_n\end{pmatrix}$. Then $$ A^\top B = M \quad\text{or}\quad a_1x_1+a_2x_2+\ldots+a_nx_n=M. $$ If $a_i=0$ for every $i \in \{1,2,\ldots,n\}$ and $M=0$ then $B$ is an arbitrary vector. If $a_i=0$ for every $i \in \{1,2,\ldots,n\}$ and $M \neq 0$ ...


1

An Eigenvector is nothing more than a vector that points to some place No, an eigenvector is a vector that points to the same place after being transformed by the matrix. As far as usefulness of eigenvectors go, it is very hard to find a mathematical concept that has more real life applications then the concept of eigenvectors: for example, the ...


1

i think you can do this with cayley-hamilton theorem which says that a matrix satisfies its characteristic equation. that is $$A^2 -(a+b)A + abI = 0 $$ or $$A^2 = (a+b)A - abI$$ we can use this to write every power of $A$ as a lineay combination of $A, I$. we will the division algorithm to find the remainder. suppose $$x^n = q(x)(x-a)(x-b) + \alpha x + ...


1

We have $$A = X \begin{bmatrix}a & 0\\0 & b\end{bmatrix} X^{-1}$$ This means $$A^n = X \begin{bmatrix}a^n & 0\\0 & b^n\end{bmatrix} X^{-1}$$ We have $$A-bI = X \begin{bmatrix}a-b & 0\\0 & 0\end{bmatrix} X^{-1} \text{ and }A-aI = X \begin{bmatrix}0 & 0\\0 & b-a\end{bmatrix} X^{-1}$$ Hence, $$\dfrac{a^n}{a-b}\left(A-bI\right) = ...


1

Think of the matrix as a function from $\Bbb R^4$ to $\Bbb R^4$. If there is line through the origin in $\Bbb R^4$ that is taken to itself by the function, then it is an eigenvector and the amount that line is stretched or shrunk is the eigenvalue. It's easier to see this in $\Bbb R^2$. If the function is a rotation of $90^\circ$ then there are no (real) ...



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