# Tag Info

6

The answer is no. Your matrix $A$ is an avatar of a companion matrix. The geometric multiplicity of an eigenvalue of such a matrix is always $1$;

4

Suppose you have a non-trivial solution of $$T^*Tf = \int_{t}^{1}\int_{0}^{s}f(y)dy ds = \lambda f(t)$$ Then $\lambda \ne 0$ because the above would give $f=0$ after differentiating a couple of times. For $\lambda \ne 0$, any solution of the above must satisfy $$\lambda f'' = -f \\ f(1)=0,\;\; f'(0)=0.$$ Any ...

4

Here's one way to find the eigenvalues: we observe that $A - I$ is a symmetric rank $2$ matrix, so it has exactly $2$ non-zero eigenvalues $\lambda_1,\lambda_2$, and it has $0$ as an eigenvalue exactly $n-1$ times. We note that $$\lambda_1 + \lambda_2 = \operatorname{trace}(A-I) = n-1\\ \lambda_1^2 + \lambda_2^2 = \operatorname{trace}((A-I)^2) = (n-1)^2 + ... 4 To expand upon the answer by loup blanc, the minimal polynomial of~A is of degree~n, namely it is X^n-a_1X^{n-1}-a_2X^{n-2}-\cdots-a_n. If some eigenvalue~\lambda had geometric multiplicity{}>1 then there would to the contrary exist a monic polynomial of degree less than~n that annihilates~A (for instance the one obtained by dividing its ... 4 Hint If A is invertible and {\bf w} = A {\bf v}, then {\bf v} = A^{-1} {\bf w}. How does this specialize if \bf v is an eigenvector of A? 3 Let \lambda=a+ib, with \lvert \lambda\rvert^2=a^2+b^2=1. The matrix$$ U=\left(\begin{array}{rr} a & -b \\ b & a\end{array}\right)\in\mathbb R^{2\times2} $$is unitary, and its eigenvalues are \lambda,\overline{\lambda}. 3 Actually A and P^{-1}AP share the same characteristic polynomial, hence they have the same eigenvalues. Note that$$\begin{align*} \det(P^{-1}AP-\lambda I_n) & = \det(P^{-1}(A-\lambda I_n)P))\\ & = \det(P^{-1})\det(A-\lambda I_n)\det(P)\\ & = \det(A-\lambda I_n). \end{align*} $$3 This is a soft question without objective answers, but I can list some cases where finding the eigenvalues (or at least one eigenvalue) is easy: Singular matrices with obvious kernels Diagonal or triangular matrices Block-diagonal matrices with easily-analyzed blocks Matrices with known spectrum that have been spectrally shifted (by adding or subtracting ... 3 Note that the eigenvalues of P are real and in [-3,3]. Let U_n be the matrix that is derived from A_n by putting the entries [1,n],[n,1] equal to 0. Then \det(A_n\pm B_n-\lambda I_n)=\det(A_n-\lambda I_n)\pm \det(U_{n-1}-\lambda I_{n-1})=p_n(\lambda)\pm q_{n-1}(\lambda). The roots of p_n are 2\cos(\frac{2\pi j}{n}),j=1,\cdots,n. The ... 3 Every matrix of rank 1 is similar to one of the Jordan forms$$ \left(\begin{array}{cc|ccc}0&1&\\&0&\\ \hline && 0 \\ &&&\ddots \\ &&&&0 \end{array}\right) \qquad\text{or}\qquad \left(\begin{array}{c|ccc}\lambda&\\\hline &0\\&&\ddots\\&&&0\end{array}\right) $$for some ... 3 The first two diagonal matrices~D commute with every matrix, so S^{-1}DS=SS^{-1}D=D for any S, so these two are each only similar to themselves. So concentrate on the final matrix~B. You can in principle compute the set \{\, S^{-1}BS\mid s\in GL(2,\Bbb R)\,\} explicitly by using the formula for the inverse and matrix multiplication. You can maybe ... 3 This is because of the Rank nullity theorem: as matrix and its transpose have the same rank, we have$$\DeclareMathOperator\rk{rank} \rk(A-\lambda I)=\rk{}{}^\mathrm t\mkern-1.5mu(A-\lambda I)\iff \dim \ker(A-\lambda I)=\dim\ker{}{}^\mathrm t\mkern-1.5mu(A-\lambda I).$$3 Let B be an n\times n-matrix and let k:=\dim N(B). Then the row-echelon form has all zeroes in its last k rows, so B^{\top} has all zeroes in its last k columns, meaning that B^{\top}e_i=0 for the last k basis vectors e_{n-k+1},\ldots,e_n. Hence \dim N(B^{\top})\geq\dim N(B) holds for all square matrices B. Then$$\dim N(B)\leq\dim ...

2

I don't fully understand the question: if you agree that $\left\{\frac{1}{\sqrt{2\pi}}, \frac{1}{\sqrt{\pi}}\cos kx, \frac{1}{\sqrt{\pi}}\sin kx\right\}$ are a basis for the function space of $2\pi$-periodic functions, it follows immediately that any function in this space can be expressed as a linear combination of these basis functions (and the $a_i$ and ...

2

There exist an invertible matrix $P$ and an a upper triangular matrix $U$ (e.g. the Jordan normal form) such that $A=PUP^{-1}$. Say that $f(x)=k_nx^n+\cdots+k_1x+k_0$. Then $$\begin{array}{rcl} f(A)=f(PUP^{-1}) & = & k_n(PUP^{-1})^n+\cdots+k_1PUP^{-1}+k_0I=\\ & = & k_nPU^nP^{-1}+\cdots+k_1PUP^{-1}+k_0I =\\ & = & ... 2 It's just that$$P^{-1}AP-\lambda I=P^{-1}(A-\lambda I)P. $$2 If Av=\lambda v and Aw=\lambda w, then for any linear combination \alpha v+\beta w we have$$ A(\alpha v+\beta w)=\alpha Av+\beta Aw=\alpha\lambda v+\beta\lambda w=\lambda(\alpha v+\beta w). $$In words, a linear combination of eigenvectors for the same eigenvalue is again an eigenvector for that eigenvalue. That said, it could happen that no such ... 2 Hint: D is a rank-1 matrix (why?) and therefore it has an eigenspace of dimension n-1 (why?). Therefore, if you can find one vector x such that Dx=\lambda x which is not in that big eigenspace, then you have found all of the eigenvalues (why?). You shouldn't have to work too hard to come up with this x. (Note: Your guesses are largely correct, ... 2 Let \lambda \in \mathbb{C} with |\lambda|=1. Then \lambda = e^{it} for t \in \mathbb{R}. Now consider the matrix$$A=\begin{pmatrix} \lambda & 0 \\ 0 & E_{n-1} \end{pmatrix} \in \mathbb{C}^{n \times n}.$$What are the eigenvalues of A? And is this unitary? 2 First off, it is well known any eigenvalue~\lambda of a unitary matrix~U has |\lambda|=1; this is so because of v\in\Bbb C^n is a corresponding eigenvector then 0\neq v^*v=v^*U^*Uv=(\lambda v)^*(\lambda v)=\lambda\overline\lambda\,(v^*v), so 1=\lambda\overline\lambda=|\lambda|^2. To conversely show that for each \lambda\in\Bbb C with ... 2 There can't be a closed form expression here (for any meaning of "closed form" that is weaker than roots of sextic polynomials). For example, try$$ H = \pmatrix{3 & 1 & 0 & 0 & 0 & 1\cr 1 & 3 & 1 & 0 & 0 & 0\cr 0 & 1 & 3 & 1 & 0 & 0\cr 0 & 0 & 1 & 3 ...

2

The algebraic multiplicity of $0$, being at least the geometric one, is at least $n-1$ in your case, more generally $n - r$ with $r$ the rank. Contrary to what you say it is not necessarily equal to this value. Now, it is always at most $n$, the dimension. Thus, in the characteristic polynomial it is either $n-1$ or $n$. Both can happen. For the $n-1$ ...

2

This should be more seen as a hint: Consider $(x_1,\ldots,x_n)=X \in \mathbb{R}^{n\times n}$, where the $x_i$ denote the column vectors of $X$. Now consider $\mathbb{R}^{n\times n}$ as a "normal" vectorspace, i.e. with column vectors, then you have $$X=\pmatrix{x_1\\ \vdots \\ x_n}.$$ Now what is meant by $AX$ in this representation? Since the columns of ...

2

For part (2): Let $𝐼_{𝑚}−𝐴𝐵$ be invertible and let’s consider the following matrix expression $(𝐼_{𝑛}−𝐵𝐴)𝐵(𝐼_{𝑚}−𝐴𝐵)^{−1}𝐴$ which you can verify simplifies to $BA$. If that is true, then what is $(𝐼_{𝑛}−𝐵𝐴)[𝐵(𝐼_{𝑚}−𝐴𝐵)^{−1}𝐴+I_{n}]$? You can easily construct a similar argument if you first let $𝐼_{𝑛}−𝐵𝐴$ be invertible.

2

As hinted in Carmichael's comment, we have that the following are equivalent: (1) $(I_n-BA)$ is invertible. (2) $1$ is not an eigenvalue of $BA$. (3) $1$ is not an eigenvalue of $AB$. (4) $(I_m-AB)$ is invertible. You've proven that $(2) \iff (3)$, so just note that $$(I-M)v \iff Mv=Iv=v$$ for any matrix $M$.

2

Just express the identity matrix as a product $\mathbf{VV^T}$ where $\mathbf{V}$ is an orthonormal matrix with first column $u/\|u\|$ and the remaining columns some basis of the perpendiculsr space of $u.$ You'll get a matrix decomposition of $H$ showing that the eigenvalues are $(-1,1,..,1),$ eigenvectors are the same as columns of $\mathbf{V}.$

1

Normally the singular values of a matrix $A$ are defined as the (positive) square roots of the eigenvalues of $A^*A$.

1

For any $n \times n$ (complex) matrix $A$, there are always invariant subspaces of $\mathbb C^n$ of all dimensions $\le n$. If $A$ is upper triangular, the $k$-dimensional subspace spanned by the first $k$ standard unit vectors is invariant. And every square matrix is similar to an upper triangular matrix.

1

For $\lambda=-3$ we consider $A-\lambda I = A-(-3)I=A+3I$ $\begin{bmatrix}5&8&16\\ 4&1&8\\ -4&-4&-11\end{bmatrix} + \begin{bmatrix}3&0&0\\0&3&0\\0&0&3\end{bmatrix} = \begin{bmatrix} 8&8&16\\4&4&8\\-4&-4&-8\end{bmatrix}$ We try to ask: what is the eigenspace for the eigenvalue $-3$? ...

1

Your '' instinct'' is correct only if we want $Ax=\lambda Bx$ for all $x$. But, if we search some $x$ for which the equation is true, than we have to solve $$\begin{bmatrix} 3&1\\ 6&-2 \end{bmatrix} \begin{bmatrix} x\\ y \end{bmatrix} = \begin{bmatrix} -1&1\\ 0&0 \end{bmatrix} \begin{bmatrix} \lambda x\\ \lambda y \end{bmatrix}$$ in ...

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