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6

The determinant is (up to sign and a factor of $6$) the volume of a tetrahedron the origin and the three column vectors as vertices. Moving one vertex further from the plane determined by the other three increases this volume. By convexity, among the farthest points is at least one vertex. Thus the result for $[-1,1]$ is the same as for $\{-1,1\}$.


5

$ (1) $ is easily seen to be true. First off, "an" identity operator is actually "the" (unique) identity operator, since any linear operator on $ V $ is uniquely determined by its action on a basis $ \{e_i\}_{i=1}^N $ of $ V $. Now, the matrix representing $ I_V $ on any basis (for instance $ \{e_i\} $) will have as its $ i $-th column $ I_V(e_i) = e_i $, ...


5

Then you are just implying zeros in a third invisible column. If you really want to call that a 3x2 matrix (that there really is no last invisible column of zeros) then I would argue that your idea of subtraction is not a closed operation, which is not desirable. You would also have non-unique additive inverses, which is also not desirable. There is nothing ...


4

There are six cases for the characteristic polynomial $\;p(x)\;$ and for the minimal one $\;m(x)\;$: $$\begin{align*}\bullet&p(x)=(x-a)(x-b)(x-c)=m(x)\;,\;a,b,c\;\;\text {different. In this case the matrices are diagonalizable:}\\ \begin{pmatrix}a&0&0\\0&b&0\\0&0&c\end{pmatrix}\\{}\\ \bullet&p(x)=(x-a)^2(x-b)=m(x)\;,\;\;a\neq ...


4

The first entry in $A^2$ comes from the first entry in $AA$, and that entry is $0\cdot 0 + 1\cdot 1 + 1 \cdot 1 = 2$. Likewise, the entry in the first row, second column of $A^2$ is $0\cdot 1 + 1\cdot 0 + 1 \cdot 1 = 1$. Perhaps you'll want to revisit matrix multiplication. In particular, Wikipedia provides this example for arbitrary $3\times 3$ ...


4

Every real $3 \times 3$ matrix, being of odd size, has a real eigenvalue since the characteristic polynomial is of odd degree. If $A^2 + I = 0$, and $A\mathbf v = \lambda \mathbf v$ with $\mathbf v \ne 0$, then $0 = (A^2 + I) \mathbf v = (\lambda^2 + 1) \mathbf v \Rightarrow \lambda^2 + 1 = 0$. Applying this notion to a real eigenvalue of $A$ leads to an ...


3

The RREF of $[A-1I]v_1=0$ is: $$\begin{bmatrix}0 & 1 & 0\\0 & 0 & 1 \\ 0 & 0 & 0\end{bmatrix}v_1 = 0$$ What if we chose: $$v_1 = (1,0,0)$$ Update If we write: $$ 0a + 1b + 0c = 0 \\ 0 a + 0b + 1 c = 0 \\ 0 a + 0 b + 0c = 0$$ What choices will actually make all three equations true and not be a zero eigenvector? What is we choose ...


3

This seems to be a bit late for an answer, but the following theorem is proven in "Linear Algebra" by Lax in chapter 10. Let $A$ and $B$ be self-adjoint, positive, $n \times n$ matrices. Then for all $0<t<1,$ \begin{align} \det(tA + (1-t)B) \geq (\det A)^{t}(\det B)^{1-t}. \end{align} Your answer follows with $t = \frac{1}{2}$.


3

This is actually very simple. The main point (seemingly missed by the OP) is that $Av_1$ and $Av_2$ must be orthogonal (this is something obvious to people familiar with proofs of SVD, because the induction decomposes into orthogonal spaces). For every $z\in{\mathbb C}$, you have $$ ||A(v_2+zv_1)||^2 \leq \sigma^2 ||v_2+zv_1||^2 \tag{1} $$ Expanding, one ...


3

Note that for $\lambda=a-b$ we have $$ A-\lambda I = \begin{bmatrix} b & \cdots & b \\ \vdots & \ddots & \vdots \\ b & \cdots & b \end{bmatrix}\tag{1} $$ The matrix in $(1)$ has rank $1$ so its nullspace has dimension $n-1$. Hence $\lambda_1=a-b$ is an eigenvalue of $A$ whose geometric multiplicity is $n-1$. Now, note that ...


3

This is the Hadamard maximum determinant problem when $n=5$. The maximum possible determinant should be $5$, which is realised by, for instance, the following matrix: $$ \pmatrix{ 1&1&0&1&0\\ 0&0&0&1&1\\ 0&0&1&1&0\\ 1&0&1&0&1\\ 0&1&1&0&1}. $$


3

When $n$ is odd use the $n \times n$ matrix for the cyclic permutation $(a_1,a_2,...,a_n) \rightarrow (a_n,a_1,...,a_{n-1})$.. Then $A^n=I$ and since $n$ is odd, $det(A)=1$ and the entries of $A$ are all $1$ or $0$. The eigenvalues are the distinct $n$'th roots of unity (it has characteristic polynomial $x^n-1$). If you allow determinant $-1$ this will ...


3

Note that your matrix can be written as $$\text{diag}(a_1,a_2,\ldots,a_n) + \begin{bmatrix} 1\\1\\1\\ \vdots\\1\end{bmatrix} \begin{bmatrix} b_1 & b_2 & \cdots & b_n\end{bmatrix}$$ This is a rank $1$ update to a diagonal matrix, whose determinant can be computed using the Sylvester determinant theorem: $$\det(I+UV^T) = \det(I+V^TU)$$ I will leave ...


3

Your matrices are similar, related by $\begin{pmatrix}0&1\\1&0\end{pmatrix}$. By inspection, what each of your matrices do is kill one basis vector while turning the other into the first one. That's intuitively similar, so to transform one to the other all we have to do is swap the basis vectors which is what ...


2

Just to supplement the other answers, here's Gaussian elimination in action: \begin{equation} \begin{bmatrix}-1 & 2 & 5\\2 & -4&a-4\\1&-2&a+1\end{bmatrix} \to \begin{bmatrix}-1 & 2 & 5\\0 & 0&a+6\\0&0&a+6\end{bmatrix} \to \begin{bmatrix}-1 & 2 & 5\\0 & 0&a+6\\0&0&0\end{bmatrix}. ...


2

If the rank of $B$ is less than $n$, then $X$ cannot exist, because the rank of $ABX$ is at most equal to the minimum rank of the factors. Therefore … Alternatively, $\operatorname{rk}A=n$ means $A$ is invertible, so you equation is equivalent to


2

$A^2$ is simply $A\cdot A$, where the $\cdot$ denotes matrix multiplication. Matrices are not multiplied elementwise (other than for addition), but as follows: The matrix element of the resulting matrix which sits on row $i$ and column $j$ is calculated by multiplying each element of $i$-th row of the first matrix with the corresponding element of the ...


2

There exists an operation of multiplication (that is NON commutative!) between matrices. First it's fundamental that the number of column of the first matrix is equal to the number of rows of the second one; so you don't have any problem with $n\times n$ matrices, like yours one. Second, the multiplication of two matrices gives a third matrix that is ...


2

Hint. In this case I wouldn't use the "smart" method of calculating the exponential, but go back to the definition via Taylor series. If $A^3=\alpha^2A$ then $A^4=\alpha^2A^2$ and $A^5=\alpha^2A^3=\alpha^4A$ and so on; therefore $$\eqalign{\exp(A)&=I+A+\frac{1}{2!}A^2+\frac{1}{3!}A^3+\frac{1}{4!}A^4+\cdots\cr ...


2

Note that if the columns of $U$ and $\hat{U}$ span the same subspace, then $\hat{U}=UM$ for some nonsingular $M$. If the columns of both $U$ and $\hat{U}$ are orthonormal, the matrix $M$ is orthogonal because $I=\hat{U}^T\hat{U}=M^TU^TUM=M^TM$. So, if $U_1=[u_1,\tilde{U}_1]$ and $\hat{U}_1=[u_1,\tilde{\hat{U_1}}]$, then $\hat{U}_1=[u_1,\tilde{U}_1]M$ for ...


2

For any square matrix with one value on the diagonal and another value everywhere else, a consistent pattern of (orthogonal but not orthonormal) eigenvectors for the $n$ by $n$ case can be read from the columns of $$ \left( \begin{array}{rrrrrrrrrr} 1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 ...


2

Let's form the Lagrangian first: $$L(x,\lambda)=x^TAx-\lambda x^Tx-\lambda=x^T(A-\lambda I)x-\lambda$$ Now, if you take derivative with respect to $x$, and set to zero you get: $$\frac{\partial}{\partial x}L(x,\lambda)=\frac{\partial}{\partial x} (x^TAx-\lambda x^Tx-\lambda)=2(A-\lambda I)x=0$$ $$\Rightarrow Ax=\lambda x$$ Therefore, $\lambda $ should ...


2

All of your work is correct for $A = PLU$. Now, we want to solve $Ax = b$ for $b_1$ and $b_2$, so we use forward and back substitution as: $$Ly = Pb, Ux = y$$ Forward substitution yields: $$Ly = \begin{bmatrix}1 & 0 & 0\\1 & 1 & 0\\ 2 & 0 & 1 \end{bmatrix}\begin{bmatrix}y_1\\y_2\\y_3 \end{bmatrix}=Pb = \begin{bmatrix}1 & 0 ...


2

I think you have the relationship between matrices and tensors wrong. The tensor $e_i\otimes e_j$ corresponds directly to a matrix (rather than to an entry of a matrix), specifically the matrix $e_ie_j^T$. I am also assuming that by tensor you mean elementary tensor product, i.e. a tensor product of the form $v\otimes w$ for some $v,w\in\mathbb R^n$. In ...


2

Most non-normal matrices whose eigenvalues are pure imaginary will have this property. (Normal means $AA^* = A^*A$.) E.g. $A = P D P^{-1}$ where $P$ is an invertible $2\times 2$ matrix that is not a multiple of a unitary matrix, and $D = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$: then $$ e^{tA} = P \begin{bmatrix} \cos(\theta) & ...


2

Proposition: Let $E=\{A\in M_n(\mathbb{C})|A\text{ has algebraic entries }\}$. Then the exponential map is injective on $E$. Proof: Assume that $e^A=e^B$. Here $A,B$ have algebraic entries ; according to "Wermuth, 2 remarks on matrix exponentials" (in free access) http://www.sciencedirect.com/science/article/pii/0024379589905545 $AB=BA$. Thus $A,B$ are ...


2

The simplest way is probably this: construct a $2\times2$ block with two of the eigenvalues, and then just use the other one. For instance $$ \begin{bmatrix}15&9&0\\-6&0&0\\0&0&2\end{bmatrix} $$ The $2\times2$ block in the upper left corner has eigenvalues $6,9$. For a more general way: You can grab any orthonormal basis $v_1$, ...


2

If $ABA^T$ is invertible then $A(BA^T(ABA^T)^{-1})=I_m$, that is, $A$ is right-invertible. Similarly $A^T$ is left-invertible with left-inverse $(ABA^T)^{-1}AB$. Write, abusing notation, $A^{-1}$ and $A^{-T}$ for the right- and left inverses respectively, and $C$ for $ABA^T$. Then $C^{-1}=(C^{-1}AB)B^{-1}(BA^TC^{-1})=A^{-T}B^{-1}A^{-1}$: the same formula ...


2

It does not matter which norm you take. It only needs to be compatible with matrix multiplication, $\|AB\|\le \|A\|\cdot \|B\|$ for all $A,B$. The problem can be solved using Neumann's series: If $\|K\|\le 1$, then $I-K$ is invertible with $$ (I-K)^{-1} = \sum_{i=0}^\infty K^i. $$ Now set $K=I-AB$, and suitably multiply the identity.


1

If I understand your notation correctly, you are right: this isn't a vector space, for the reasons you gave. For example, $A = \left(\begin{array}{cc}1&0\\0&-1\end{array}\right)$ and $B = \left(\begin{array}{cc}0&0\\0&1\end{array}\right)$ are both in that set, but $A+B = \left(\begin{array}{cc}1&0\\0&0\end{array}\right)$ is not in ...



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