# Tag Info

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 $$\left( \begin{array}{rr} 1 & 0 \\ 0 & 0 \end{array} \right)$$

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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 ...

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$\pmatrix{a&a&-1&1\\ 1&-1&1&a\\ -1&1&a&1\\ }$ $\implies\pmatrix{1&1&-\frac1a&\frac1a\\ -1&1&-1&-a\\ -1&1&a&1\\ }$ 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 $(AB)^T = B^TA^T$.

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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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