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1

This question can be generalized as follow: Aim: Find the transformation matrix $M$ such that it projects every point on xy-plane into the line $l_1$ where $l_1$ has equation $y=mx$. Suppose $A=(a,b)$ is an arbitrary point on the xy-plane. Denote the line joining the point $A$ and perpendicular to the line $y=mx$ as $l_2$. Then $l_2$ has gradient ...


0

suppose $A=(a,b)$, because you want the project of $A$ on the line $y=4x$, $A$ should be on a line which is perpendiculaire to the known line. So the tangent of the new line is $-1/4$. Then the points on the new line could be expressed as $A+t(1,-1/4)=(a+t,b-t/4)$. The project of $A$ on the line is denoted by $A'$, then there should be some $t$ satisfying ...


0

No. For example with $R=\Bbb{Z}$ we can use $$ A=\left(\begin{array}{cc}2&0\\0&2\end{array}\right)\qquad\text{and}\qquad B=\left(\begin{array}{cc}4&0\\0&1\end{array}\right). $$ Here for all $X\in GL_2(\Bbb{Z})$ all the entries of $XA$ are even, so we never get $XA=B$. Yet $\det A=\det B=4$. In general the answer revolves around the concept ...


1

No. You have described all the real symmetric matrices with nonzero determinant. The others are usually called semidefinite, for example $$ \left( \begin{array}{rr} 1 & 0 \\ 0 & 0 \end{array} \right) $$


0

Yes. By the way, check out Wolfram's definition of Indefinite Matrix http://mathworld.wolfram.com/IndefiniteMatrix.html


2

The determinant is the same polynomial in the matrix entries no matter which field (or commutative ring) the entries come from. So what you're doing is right -- you can think of it either as doing the calculations in $\mathbb Z_5$, or as computing the determinant over $\mathbb Z$ and reducing modulo 5 at the very end.


2

Hint: Pick two rows (or two columns). Are they linearly independent?


0

A "standard" generator matrix of a $[n,k]$ code usually means that we seek a $k\times n$ matrix $G$ of the form $$G = \left[I_{k\times k}\ P_{k\times (n-k)}\right]$$ where $I_{k\times k}$ denotes a $k\times k$ identity matrix. For a $[7,6,2]$ code, $P_{6\times 1}$ is just a column of $6$ bits, and I leave it to you to figure what the six bits must be in ...


0

$\pmatrix{a&a&-1&1\\ 1&-1&1&a\\ -1&1&a&1\\ }$ $\implies\pmatrix{1&1&-\frac1a&\frac1a\\ -1&1&-1&-a\\ -1&1&a&1\\ }$ by $R_1\to\frac1aR_1, R_2\to-R_2$ $\implies \pmatrix{1&1&-\frac1a&\frac1a\\ 0&2&-1-\frac1a&-a+\frac1a\\ 0&2&a-\frac1a&1+\frac1a\\ }$ by ...


0

For $\ a=1$ you have: $$\begin{cases}x=0 \\ y=0 \end{cases}$$ For $\ a≠1$ $$\begin{cases}(a-1)x+2y=0 \\ 2x+(a-1)y=0 \end{cases}\implies \begin{cases}(a-1)x+2y=0 \\ x=\frac{1-a}{2}y \end{cases}\implies \begin{cases}-\frac{(a-1)^2}{2}y+2y=0 \\ x=\frac{1-a}{2}y\end{cases}$$ $$\implies \begin{cases}y \left(2-\frac{(a-1)^2}{2}\right)=0 \\ ...


0

It is not possible to find this matrix $B_{n\times m}$, $m<n$, for any real matrix $A_{n\times n}$ if $\lambda$ is not real. Let $A_1,\ldots,A_n$ be the columns of $A$. Let $\lambda=a+bi$ with $b\neq 0$. Let $C_1,\dots,C_n$ be the columns of $\lambda Id-A$. Thus $C_j=(ae_j-A_j)+i(be_j)$, where $\{e_1,\ldots,e_n\}$ is the canonical basis of ...


4

The set on invertible upper triangular matrices is actually closed in $GL(n,\Bbb R)$, since it is defined by the vanishing of a bunch of matrix entries (which entries are continuous functions of the matrix). If it were also open, it would be a union of connected components. But $GL(n,\Bbb R)$ has only two connected components (determined by the sign of the ...


1

Let me write the matrix in question in the form $L:=D-A$ (instead of $D+A$ to avoid alternating signs). Let $D$ be nonsingular and $\|D^{-1}A\|=:\epsilon<1$ for some operator matrix norm. We can write $$ L=D(I-D^{-1}A). $$ Using the Neumann series and with $B:=D^{-1}-D^{-1}AD^{-1}$, we have $$ ...


0

If, e.g., $A$ is lower triangular, you can implement the inversion algorithm by overwriting the storage occupied by $A$. The inversion is essentially solving a system $AB=I$ with multiple right-hand sides composed of the columns of the identity matrix. Say, we want to compute the $k$-th column of $B$, that is, to solve the system $$ Ab_k=e_k, $$ where $e_k$ ...


1

Ok, in English it is symmetric matrix. There must be some additional conditions you didn't tell. But if they are fulfilled: What did you do yourself? What is the transpose of $X^tX$? ( Wikipedia is your friend, if you don't know) What is for a vector $a$ then $Xa$ and what is $a^tX^t$ (if the not by you provided conditions are fulfilled)? From this you ...


3

Presumably, the author meant positive semidefinite, or specified something about the rank of $X$. Hint: Note that a matrix $A$ is positive semidefinite iff $v^TAv \geq 0$ for all vectors $v$. Note that $v^TX^TXv = (Xv)^T(Xv)$. As for symmetry: note that $(AB)^T = B^TA^T$.


1

Matrix similarity: $\DeclareMathOperator{\rank}{rank}$ We say that two similar matrices $A,B$ are similar if $B = SAS^{-1}$ for some invertible matrix $S$. In order to show that $\rank(A) = \rank(B)$, it suffices to show that $\rank(AS) = \rank(SA) = \rank(A)$ for any invertible matrix $S$. To prove that $\rank (A) = \rank(SA)$: let $A$ have columns ...


1

If $n$ vectors $v_1,..v_n$ are linearly independent(dependent) then for non singular matrix $P$ the vectors $Pv_1,...Pv_n$ also will be independent(dependent). From this fact and from the definition of the rank as a number of linearly independent columns (rows) we immediately can conclude that similar matrix have the same rank. The second fact follows ...


2

One purely matrix-based definition of rank is decomposition rank: the rank of an $n\times m$-matrix is the smallest integer $r$ such that the matrix can be decomposed as product of an $n\times r$ and a $r\times m$ matrix. It is now obvious that the rank of $AB$ cannot be larger than the rank of $A$, or than the rank of $B$ (a decomposition of $A$ or of $B$ ...


1

I will assume in the following that $A$ is a hermitian matrix. Then the matrix $A' = U ^{\dagger} A U$ is again hermitian. Let set of possible diagonals value $ (d_1, \ldots d_n)$ achieved by the matrices of form $A' = ^{\dagger} A U$ forms a polytope in the hyperplane $\sum d_i = \text{trace} A$. This is an important theorem of Horn. The closest point ...


2

As $A$ and $B$ positive definite they have positive definite square roots: $$ A=A_1^2,\,\, B=B_1^2. $$ Clearly$^*$, $$ A-B\ge 0 \Longleftrightarrow B^{-1}_1AB_1^{-1} \ge I, $$ where $I$ is the unit matrix. Also$^{**}$, $$ B^{-1}_1AB_1^{-1} \ge I\Longleftrightarrow I \ge B_1A^{-1}B_1 \Longleftrightarrow B^{-1} \ge A^{-1}. $$ $^*$More specifically, if ...


1

The simple Theorem to remember is that given a basis $\{\alpha_1, \alpha_2, ..., \alpha_n\}$ of $V$ and any $n$ vectors $\{\beta_1, \beta_2,.., \beta_n, \}$ in $W$ there is exactly one Linear Transformation such that $ T(\alpha_i) = \beta_i$. So you have two simple tasks: Find a basis $\{v_1, v_2\} $for $\ker T$ and let $T(v_1) = T(v_2) = \underline {0}$ ...


2

For number you have $$ \frac{1}{x^{-1} + y^{-1}} = \frac{xy}{y+x}$$ For matrices, let $Z = X^{-1} + Y^{-1}$, I want to show that $X(X+Y)^{-1}Y Z = I$. $$ X(X+Y)^{-1}Y ( X^{-1} + Y^{-1} ) = X(X+Y)^{-1}YX^{-1} + X(X+Y)^{-1}YY^{-1} = $$ $$ = X(X+Y)^{-1}(YX^{-1} + I) = X(X+Y)^{-1}(YX^{-1} + I)XX^{-1} = $$ $$ = X(X+Y)^{-1}(Y + X)X^{-1}) = XX^{-1} = I$$


3

We have $$(X(X+Y)^{-1}Y)^{-1}=Y^{-1}(X+Y)X^{-1}=Y^{-1}XX^{-1}+Y^{-1}YX^{-1}\\=Y^{-1}+X^{-1} $$ and the result follows.


1

a) First note that if we have a sum $$f(a)=\sum_{k=1}^N \lambda_k \exp(iax_k)$$ with the $x_k\in \mathbb{R}$ distincts, then this function is zero on $\mathbb{R}$ if and only if all $\lambda_k$ are zero. To see this, use induction on $N$, and if $f(a)=0$ for all $a$, compute the derivative of $g(a)=f(a)\exp(-iax_N)$. b) Now prove your assertion by ...


0

Certainly it is possible. The condition is equivalent to $I = B A (B A)^t$, i.e. that $BA$ is an orthogonal matrix. No need for $BA = AB$, or for $AB$ to be orthogonal.


3

Note that if $A = \exp(B),$ then $B$ must commute with $A$, hence must also be diagonal. Let $B = [b_{ij}]$. Then $\exp(b_{11}) =-2$ and $\exp(b_{22}) = -1.$ Therefore $b_{11}$ and $b_{22}$ are not real.


0

Perhaps you would be interested in this link? By this method, we know that the matrix would have to be complex to cope with the logarithms of negative numbers.


3

Maximal dimension of a commutative subalgebra of $n$ by $n$ matrices is $$ 1 + \left\lfloor \frac{n^2}{4} \right\rfloor. $$ In even dimension $2m,$ this is realized by $$ \left( \begin{array}{rr} \alpha I & A \\ 0 & \alpha I \end{array} \right), $$ where $A$ is any $m$ by $m$ matrix and $I$ is the $m$ by $m$ identity. This is a theorem of ...


1

The system is $$ \begin{pmatrix} x \\ y \end{pmatrix} = \begin{bmatrix} \cos \theta & - \sin\theta \\ \sin \theta & \cos\theta \end{bmatrix} \begin{pmatrix} a \\ b \end{pmatrix} $$ If the 2×2 matrix is a rotation, when you invert it you will get the inverse rotation. So you either do it the long way (with 2×2 matrix inversion) or the short way of ...


2

Hint: What is the opposite operation of rotating a vector by $\theta$ in the anti-clockwise direction?


3

Up to conjugacy, there are three maximal abelian subrings in the ring of two by two matrices with real entries: $$\left\{ \left( \begin{matrix} a & 0 \\ 0 & b \end{matrix} \right) \right\}, \ \left\{ \left( \begin{matrix} a & b \\ -b & a \end{matrix} \right) \right\}, \ \text{and} \ \left\{ \left( \begin{matrix} a & b \\ 0 & a ...


1

I have seen this notation used (sometimes with "()" rather than "[]"), and I would think that most people reading any linear algebra text would know what you mean. If you want to be more explicit, you could right something like $X = [x_{ij}]_{i,j = 1}^n$ to mean "let $X$ denote the matrix whose entries are $x_{ij}$ where $i$ and $j$ go from $1$ to $n$".


5

Let me consider $W:=-V$; we have $$ \det(V)=(-1)^n\det(W). $$ As already noted in the comments, $W=D-(ve^T+ev^T)$, where $v:=[v_1,\ldots,v_n]^T$, $e:=[1,\ldots,1]^T$, and $D:=\mathrm{diag}(\beta_i v_i)_{i=1}^n$. Since $D>0$, we can take $$ D^{-1/2}WD^{-1/2}=I-(\tilde{v}\tilde{e}^T+\tilde{e}\tilde{v}^T), $$ where $\tilde{v}:=D^{-1/2}v$, ...


2

By investigating the structure of the result I come to the nice general formula $$ \left|V\right| = (-1)^n \left(1-\sum _{i=1}^n \left(\frac{2}{\beta _i}+\sum _{j=1}^{i-1} \frac{\left(v_i-v_j\right){}^2}{v_i v_j \beta _i \beta _j}\right)\right) \prod _{k=1}^n v_k \beta _k $$ If you have Mathematica you can use the following code to check the result ...


2

For 1, multiply the second equation by $3$ and subtract the first: $$ A=6A^t-3I_2 $$ Now transpose: $A^t=6A-3I_2$, so $$ A=6(6A-3I_2)-3I_2 $$ or $$ 35A=21I_2 $$ Once you know $A$, it's easy to compute $B$ from the first equation. For 2, transpose the second equation and eliminate $B$, then do similarly.


0

take elements of $A$ as $x_1,x_2,x_3,x_4$ and $B$ as $y_1,y_2,y_3,y_4$ substitute them in the equaions and also find $A^T$ after simplifying then compare the rows and columns. you will get equations solve it and get the results of $x_1,x_2,x_3,x_4,y_1,y_2,y_3,y_4$ and there you find $A$ and $B$


1

$[A^T,B^T]^T$ is what you want. Also, $[A,B]$ only makes sense if you know that $A$ and $B$ share the same number of rows.


7

Let's call your matrix $$A = \left( \begin{array}{cc} 2 & 3 \\ 1 & 4 \end{array} \right)$$ We want a matrix $X_{2\times 2} = \begin{pmatrix} a & b\\ c&d\end{pmatrix}$ such that $AX = XA$. $$AX = \begin{pmatrix} 2a + 3c & 2b+3d\\ a + 4c&b+4d\end{pmatrix}$$ $$XA = \begin{pmatrix} 2a + b&3a + 4b\\2c+d & 3c+4d\end{pmatrix}$$ ...


2

If you write down the unknown matrix as $$ \begin{bmatrix} a & b \\ c & d \end{bmatrix} $$ Then you want $$ \begin{bmatrix} a & b \\ c & d \end{bmatrix} \cdot \begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix} = \begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix} \begin{bmatrix} a & b \\ c & d \end{bmatrix} $$ Write out the left ...


4

You need $$\left(\begin{array}{cc}a&b\\c&d\end{array}\right) \left(\begin{array}{cc}2&3\\1&4\end{array}\right)= \left(\begin{array}{cc}2&3\\1&4\end{array}\right)\left(\begin{array}{cc}a&b\\c&d\end{array}\right)$$ Multiply out the matrices; that will give you four equations that connect $a,b,c$ and $d$. Then solve those ...


4

You can just use $$A = \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}$$ $$B = \begin{pmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{pmatrix}$$ and then solve the system of equations for $a_{ij}, b_{ij}$. For each matrix-equation you will get 4 scalar equations (one for each entry). You can also try to simplify the ...


2

Since $SDT$ is symmetric, we have $SDT=T^TDS^T$ and hence $DTS^{-T}=S^{-1}T^TD$. Denoting $X=TS^{-T}$, we have $$\tag{1} DX=X^TD. $$ Note that $X$ is upper triangular because $T$ and $S^{-T}$ are upper triangular. However, (1) implies that $X$ is also lower triangular and consequently, $X$ is diagonal. It means that $X=TS^{-T}=C$, where $C$ is a non-singular ...


0

No, $ST$ is not necessarily symmetric. For example $$\left(\begin{array}{cc}1&4\\1&2\end{array}\right)\left(\begin{array}{cc}1&0\\0&\frac{1}{2}\end{array}\right)\left(\begin{array}{cc}-1&2\\1&-1\end{array}\right)=I,$$ and if you omit the middle matrix, which is diagonal, you'll get a non-symmetric matrix.


2

Let me summarize my comments in an answer. I don't have answers to all the questions but I don't expect (myself) to really come up with any more, so here goes. First, I am mostly thinking about this space as the algebraic variety over $\mathbb{R}$ (complex conjugation is not $\mathbb{C}$-scalar) cut out by the $n^2$ (in the $\mathbb{R}$ case) or $2n^2$ (in ...


0

The OP's question would have been well-phrased had he specified that the matrix $A$ is symmetric i.e. $A=A^\top$, in which case ${\rm colspan}(A)={\rm rowspan}(A^\top)={\rm rowspan}(A)$. Now, consider the definition of ${\rm null}(A)$ as the space of all vectors $\mathbf{v}$ such that $A\mathbf{v}=\mathbf{0}$. Letting $\mathbf{a}_1,\ldots,\mathbf{a}_n$ be ...


1

As explained in the comments, the problem is one of extending the two matrices in such a way that 1) the same non-commutativity can be easily verified and 2) the matrices are non-singular. A standard recipe is to extend by ones along the diagonal and zeros elsewhere. Why is this "standard"? The way I think about it is in terms of linear transformations. ...


0

$GL_n(F)\cong Aut(F^n)\,\, n\geq 3.$ where $F$ is an arbitrary field (not necessarily of characteristic zero). Let $f(x_1,x_2\cdots,x_n)=(x_2,x_1,\cdots,x_n)$ and $g(x_1,x_2,\cdots,x_n)=(x_3,x_2,x_1,\cdots,x_n)$ Clearly $f,g\in Aut(F^n)$. Notice that $f\circ g(1,0,0,\cdots,0)=f(0,0,1,0\cdots,0)=(0,0,1,0,\cdots,0)$ but $g\circ ...


3

Note that $M$ = $\begin{bmatrix}m_1&m_2&\cdots&m_n\\m_1&m_2&\cdots&m_n\\\vdots&\vdots&\ddots&\vdots\\m_1&m_2&\cdots&m_n\end{bmatrix} + \begin{bmatrix}m_1&m_1&\cdots&m_1\\m_2&m_2&\cdots&m_2\\\vdots&\vdots&\ddots&\vdots\\m_n&m_n&\cdots&m_n\end{bmatrix}$. Hence, ...


1

If we assume that $A$ is normal and non-singular (there are no singular, non-zero solutions), then this becomes an equation on eigenvalues: $$ |\lambda_1|^2 + |\lambda_2|^2 = 2|\lambda_1|^2 |\lambda_2|^2 \implies\\ 2|\lambda_1|^2 |\lambda_2|^2 - |\lambda_1|^2 - |\lambda_2|^2 = 0 \implies\\ \frac 12(2 |\lambda_1|^2 - 1)(2|\lambda_2|^2 - 1) - \frac 12 = ...



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