New answers tagged linear-algebra
1
I am not sure what exactly you are asking. If you ask why $\det \prod_{k=1}^N M_k=1$ when $\det M_1,\ldots,\det M_N=1$, this is because the determinant function is multiplicative (i.e. $\det AB=(\det A)(\det B)$). If you are asking why a sympletic matrix has determinant $1$, this has been explained in the related Wikipedia article. If you are asking why ...
1
Use $det(AB...Z)=det(A)det(B)...det(Z)$ and the fact that $det(M_k)=1$.
0
Hint: Use elementary row operations to bring $C$ to its reduced row echelon form $R$, then use elementary column operations to bring $R$ to its reduced column echelon form $C'$.
0
Which definition of $A>0$ do you use?
Let $V$ be arbitrary inner product space. We say a linear map $A$ is pos.definite, if $A=A^*$ and for $x\in V$, $x\neq 0$ we have $\langle x,Ax\rangle>0$. Since $(AB)^*=B^*A^*$, $I=A^*(A^{-1})^*=A(A^{-1})^*$, so the unicity of the inverse yields $(A^{-1})^*=A^{-1}$. Furthermore,
$$
\langle x, A^{-1} ...
3
There is a theorem, that says that self-adjoint maps are allways diagonalizable, so there an eigenbasis and you are done.
Let's sketch the proof: Let $\lambda\in \mathbb C$ be an eigenvalue of $T$, $E_\lambda := \ker(\lambda- T)$ the eigenspace. Then $E_\lambda$ is $T$-invariant, and by self-adjointness, its orthogonal complement $E_\lambda^\bot$ is also: ...
0
First, a notational issue. The vector you call $|w\rangle$ "double u" is actually called "omega" i.e. $|\omega\rangle$ (backslash-omega is the $\rm\TeX$ command).
The vector $|s'\rangle$ is determined as the only (up to an arbitrary overall normalization including the phase) linear combination of $|\omega\rangle$ and $|s\rangle$ – note that you use the ...
0
$2:True:$ $FF^n=0\\\implies\text{Im}(F^n)\subset\text{Kernel}(F)\subset\text{Kernel}(F^n)\\\implies\text{Rank}(F^n)\le\text{Nullity}(F^n)$
$1,3,4: False:$ $F:\mathbb R^3\to\mathbb R^3$ given by $F=\begin{pmatrix}0&0&1\\0&0&0\\0&0&0\end{pmatrix}$
0
I'd recommend you check out the Mathematics section here for free online video courses offered by some of the top institutions.
7
Let $C = AB - BA$ and assume that $(C - I)^n = 0$.
Then the minimal polynomial $m(x)$ of $C$ divides $(x - 1)^{n}$. Thus $m(x) = (x-1)^{k}$ for some $1 \leq k \leq n$. This shows that the only eigenvalue of $C$ (considered in $\Bbb{C}$) is $1$, hence the characteristic polynomial of $C$ is $(x-1)^{n}$. In particular, $\operatorname{tr}(C) = n > 0$. This ...
2
I explain the relationship between Jordan and rational forms in these recently written notes. Here are the main points:
1) Jordan form is only available when a transformation is split, i.e., when its minimal polynomial (equivalently, its characteristic polynomial) is a product of linear factors over the ground field.
2) In this case, the rational ...
4
Let's go step by step. We want an equation of the following form:
...
3
As @Julien pointed out, every square matrix admits a $PLU$ decomposition, where $P$ is a permutation matrix. We have: $A = P \cdot L \cdot U$, such that:
$A=\begin{bmatrix}1 & 2 & 3 & 4 \\5 & 6 & 7 & 8\\1 & -1 & 2 & 3 \\2 & 1 & 1 &2 \end{bmatrix}= \begin{bmatrix} 1 & 0 & 0 & 0\\0 & 0 & 1 ...
1
No Conclusion
Possible eigenvalues of a non-singular real skew-symmetric matrix are purely imaginary. Consequently such matrices have no eigenvalue in $\mathbb R$ i.e. the number of eigenvalues is $0.$
The roots of the characteristic polynomial of a $n\times n$ real symmetric matrices are all real whence the number of eigenvalues is $n.$
You may form a ...
4
Since you observed that $P$ is annihilated by $X^n-1$, it only remains to check that no nonzero polynomial of degree $\leq n-1$ annihilates $P$ to conclude that the minimimal polynomial is $X^n-1$. Since the latter has degree $n$ and divides the characteristic polynomial which is also monic of degree $n$, it will also follow that the characteristic ...
2
To solve
$$
x'=Ax+b
$$
use
$$
x=e^{At}x_0-A^{-1}b
$$
where $e^{At}x_0$ is the solution to $x'=Ax$.
Often times it is useful to look at the $1$-dimensional analog:
$$
x'=ax+b
$$
where the solution is
$$
x=x_0e^{at}-b/a
$$
I just put that solution into matrix form and checked it out to make sure it worked. Low tech math.
2
If there exists a vector $v$ such that $Av=b$ then consider the variable $y=x-v$. One finds $y'=Ay$.
If there is no such $v$ then $A$ is not invertible. If $A$ is still diagonalizable, however, then the tricky part of $b$ is the bit lying in the kernel of the matrix. Thus write $b=b_0+c$ where $Ab_0=0$ but there is some $Av=c$ solution. Try computing ...
0
One common generalization to the case of square matrices is to define $$A \ge 0 :\Leftrightarrow A \; \textbf{is} \; \textbf{positive} \; \textbf{semidefinite},$$ that is, $v^T A v \ge 0$ for every $v \in \mathbb{R}^n$. (assuming you're working with real matrices).
Some common inequalities carry over immediately, appropriately modulated: for example, $$x^2 ...
1
when you consider $ \{ (1,-1),(0,1)\}$ as base of domain's T and you catch $\begin{bmatrix}1 & 2\\ 0& 1\\ -1 & -1\end{bmatrix}$ then $(-1,-1)=1(1,-1)+0(1,0)$ so :
$T(1,-1)=\begin{bmatrix}1 & 2\\ 0& 1\\ -1 & -1\end{bmatrix}\begin{bmatrix}1 \\ 0\end{bmatrix}=(1,0,-1)$
0
Look at the laplace expansion of the determinant of $A+I$ (http://en.wikipedia.org/wiki/Laplace_expansion)
Your matrix has the form $$
A = d\left(
\begin{matrix}
k_{1,1} &\ldots &k_{1,n} \\
\vdots &\ddots &\vdots \\
k_{n,1} &\ldots &k_{n,n}
\end{matrix}\right)
$$
and by laplace expanding the determinant using the ...
5
Hint: Can $-1$ be an eigenvalue for $A$?
Alternatively: What is the determinant of $A+I$ modulo $d$?
1
Hints: First,
$$
\begin{align}
(3v+w)\times(v-2w)
&=-7v\times w
\end{align}
$$
and
$$
\|-7v\times w\|=7\|v\|\|w\|\left|\sin\left(\frac{2\pi}{3}\right)\right|
$$
For $\langle w\times 5v, v-3w\rangle$, note that $w\times v$ is perpendicular to both $v$ and $w$.
2
Hints: a real symmetric matrix is diagonalizable (via an orthogonal matrix, but you don't even need that here) in $M_n(\mathbb{R})$ and the condition $A^k=Id$ implies that the eigenvalues of $A$, which are all real, belong to $\{\pm 1\}$. Finally, note that the result is true.
1
If $A \succ 0$, $S$ is positive semidefinite iff the block matrix is positive semidefinite.
1
If $B$ commutes with $A$, it leaves the eigenspaces of $A$ invariant. In this case, these are the one-dimensional subspaces generated by the vectors of the canonical basis $\{e_i\;;\; i=1,2,3\}$. So $Be_i=\lambda_ie_i$ for $i=1,2,3$. That is $B$ is diagonal. The converse is clear.
Conclusion: the commutant of $A$ in $M_3(\mathbb{R})$ is the subalgebra of ...
0
A matrix $B$ commutes with $A$ iff $AB=BA$ iff $B=ABA^{-1}$; fill the entries of $B$ with unknowns, write out this equation, and then find an equivalent system of relations for the unknowns:
$$\begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} = \begin{pmatrix}-1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & ...
1
Seems about right to me. The point is that a matrix is unitary if and only if its columns form an orthonormal basis with respect to the complex Euclidean inner product. In fact, if $U = [v_1 | \cdots | v_n]$ then $(UU^\dagger)_{ij} = \langle v_i, v_j \rangle$ (or $U^\dagger U$ depending on your convention). Thus your question really boils down to: can any ...
3
if $rank(A^*A)=rank(A)$ it means $rank(A)=n-dim(null(A))$ and $rank(A^*A)=n-dim(null(A^*A))$ so $dim( null(A))=dim(null(A^*A))$ if exist $x$ such that $Ax\ne 0$ and $A^*Ax=0$ then $x\in null(A^*A)$but but x not belong to $null(A)$ and since $null(A)\subset rank(A^*A)$ so $$dim( null(A))\ne dim(null(A^*A))$$contradiction!
on the other hand :
we know ...
0
Perhaps this is slightly off topic but it's worth explaining slightly more intuitively why you can't divide by a vector.
The key idea is to think about information (or more precisely numbers of constraints/equations). A dot product takes two vectors, both with $n$ numbers in them, and combines them in a particular way (related to projecting then onto each ...
1
I'd take a cross product, say, $(P-Q)\times(R-Q)$ to get a vector orthogonal to the plane, normalize that to get a normal vector, then form the scalar product of $Z-Q$ with that normal vector and take the absolute value to get the distance of $Z$ from the plane.
5
Note that
$$\mathrm{Tr}\left[(A-A^T)^2\right]=-2\sum_{1\le i\le j\le 3}(a_{ij}-a_{ji})^2.$$
Therefore the identity can be rewritten as
$$2(1-\mathrm{Tr}\,A)^2-\mathrm{Tr}\left[(A-A^T)^2\right]=8.$$
Since $\mathrm{Tr}\,AA^T=\mathrm{Tr}\,1=3$, this is equivalent to
$$\left(\mathrm{Tr}\,A\right)^2-\mathrm{Tr}\,A^2-2\mathrm{Tr}\,A=0.$$
The left side of the last ...
3
Hint: what would it mean for a vector to be orthogonal to itself?
If the problem says "every other vector" as in the title (but not the quoted text) you could instead consider what it means for a given vector $\vec{u}$ to be orthogonal to multiples of itself, e.g. to $-\vec{u}$.
1
We know that $G=GL_n(\mathbb{R})$ is generated by elementary matrices. Therefore $A$ is in $Z(G)$ if and only it commutes with all elementary matrices. By checking this condition, you will see that $A$ has to be diagonal.
7
The dot product $\cdot$ is an unusual type of multiplication. It takes two vectors in and produces a scalar out. Imagine a mommy elephant and a daddy elephant giving birth to a giraffe.
There's no such thing as division in this context. You need to do the proof by looking at components of the vectors. $u=(u_1,u_2,\ldots, u_n), v=(v_1, v_2,\ldots, v_n)$, ...
1
Let $A \in Z(GL_n(\mathbb R))$.
Let $B$ be defined as $b_{ii}=i$ and $b_{ij}=0$ if $i\neq j$ (or any other diagonal matrix with pairwise distinct entries).
Then since $AB=BA$, for all $i \neq j$ we have
$$\sum_{k=1}^n a_{ik}b_{kj}=\sum_{k=1}^n b_{ik}a_{kj} \Rightarrow a_{ij}b_{jj}=b_{ii}a_{ij}\,.$$
Since $i \neq j$ we have $b_{ii} \neq b_{jj} ...
0
Well, I found the answer. It is easy to prove that $x_v \le \lambda x_u$ simply by writing down $Ax=\lambda x$. Now, since $A$ is irreducible, there must exist a power $k \le m$ such that state $u$ is connected to state $v$ by a path of length $k$. $A^k$ represents a graph with paths of length $k$ between the states of the $A$-graph. This matrix has ...
0
$V_\mathbb{R} = V \otimes_\mathbb{Q} \mathbb{R}$ is a $\mathbb{R}$-vector space in "the natural way". You do not need to talk about basis elements of neither $V$ nor $\mathbb{R}$ to define the structure.
Any element of of $V_\mathbb{R}$ looks like a sum of terms $v \otimes r$, where $v \in V$ and $r \in \mathbb{R}$. To define a $\mathbb{R}$-vector space ...
1
Since that $(1,-1)$ is the first vector of $\beta$ it has coordinates $(1,0)$.
When you use matrices to apply the linear transformation you are using the coordinates of the vector on a fixed basis.
So you should write $$\begin{bmatrix}1 & 2\\ 0& 1\\ -1 & -1\end{bmatrix}\begin{bmatrix}1 \\ 0\end{bmatrix}.$$
If $\{v_1,\ldots,v_n\}$ is a basis ...
1
It suffices to prove that if $a v = 0_E$ (I take there's a misprint in you formula, and I'm simplifying notation a bit), and $a \ne 0_K$, then $v = 0_E$.
Now if $a \ne 0_K$, then $a$ has an inverse $a^{-1}$ in the field $K$.
Thus
$$
0_E = a^{-1} \cdot 0_E = a^{-1} (a v) = (a^{-1} a) v = 1 v = v.
$$
1
Given 3 vertices, these vertices define 2 vectors; the unknown endpoint corresponds to the sum of these 2 vectors.
Example: $(1,1)$ and $(1,3)$ form the vector $(0,2)$. $(1,1)$ and $(4,2)$ form the vector $(3,1)$. Sum these vectors to get $(3,3)$. Then add the vertex $(1,1)$ back to get $(4,4)$ as the formerly unknown corner.
2
The three possibilities come from adding the coordinates of two points and subtracting the third. So $(4,0)=(4,2)+(1,1)-(1,3)$. There are three choices of which point to subtract. You are translating the point you subtract to the origin, then adding the vectors to the other two points to find the opposite one, then translating back. So you could look at ...
0
Of course! $\mathbb{Q}$ is a field, therefore if there is solution to linear system, the solutions will be rational.
You can check it by writing the explicit formulas for the solution -- for example, the ones you provided as help or Cramer's rule -- and notice that the formulas involves only additions and multiplications of rationals.
So if your system has ...
1
Hints (if I guessed correctly your notation and part of your problem):
$$\begin{pmatrix}5&3&7&0\\
2&\!\!-4&6&5\end{pmatrix}\begin{pmatrix}x_1\\x_2\\x_3\\x_4\end{pmatrix}=\binom 50\iff\begin{cases}5x_1+3x_2+7x_3=5\\{}\\2x_1-4x_2+6x_3+5x_4=0\end{cases}$$
Two particular solutions to this system are
...
4
Let $\mathcal{A} = \cup_{i=1}^n A_i$ and $\chi_i : \mathcal{A} \to \mathbb{N}$ be the indicator function for $A_i, i = 1,\ldots n$.
For any $n$ real numbers $x_1, \ldots, x_n$, not all zero. Consider the function $f: \mathcal{A} \to \mathbb{R}$ defined by:
$$f(a) = \sum_{i=1}^{n} x_i \chi_i(a)$$
We have:
$$
\sum_{i=1}^{n}\sum_{j=1}^{n} a_{ij} x_i x_j
= ...
2
A fun problem. I think that the following approach is natural (at least it is the first that occurred to me, YMMV).
Let us consider the union
$$
A=\bigcup_{i=1}^nA_i.
$$
I will work in the space $F_A$ of real valued functions from $A$ to $\mathbb{R}$.
If you list the elements of $A$ like
$$
A=\{a_1,a_2,\ldots,a_m\},
$$
you can identify the space $F_A$ with ...
0
Hint:
Claim: A simetric matrix $X=(x_{ij})_{VV}$ (here $V=\{1,2,\ldots, k, \ldots, n\}$ ) is positive semidefinite if and only if all of its principal minors $det(x_{ij})_{UU}$ (here $\emptyset \neq U\subset V$) are nonnegative.
Claim: A simetric matrix $\begin{pmatrix}A& C^T\\ C& D\end{pmatrix}$ is positive semidefinite if and only if $D$ and ...
0
Recall that adding a multiple or subtracting a multiple of one row does not change the value of the determinant, see, for example ProofWiki.
Using this fact and Laplace expansion you get
$$|A_4|=
\begin{vmatrix}
1& 1& 1& 1\\
1& 2& 2& 2\\
1& 2& 3& 3\\
1& 2& 3& 4
\end{vmatrix}=
...
3
Let's prove that a linear function is continuous. Recalling the definition:
A function $f:\Bbb{R}^n\to\Bbb{R}^k$ is continuous iff for each $x\in\Bbb{R}^n$ and for each $\epsilon>0$ there exists a $\delta>0$ such that for all $y\in\Bbb{R}^n$ with $||y-x||<\delta$ we have $||f(y)-f(x)||<\epsilon$.
Now if $f$ is linear, then it takes the form ...
0
$\def\rk{\mathop{\mathrm{rank}}}$We have, as $\rk A = \rk A^t$ holds for every matrix, that
\begin{align*}
\dim U &= \rk \begin{pmatrix} a_1 & b_1 & c_1\\ a_2 & b_2 & c_2\\ a_3 & b_3 & c_3 \end{pmatrix}\\
&= \rk \begin{pmatrix} a_1 & b_1 & c_1\\ a_2 & b_2 & c_2\\ a_3 & b_3 & c_3 ...
1
$\def\Mat{\mathrm{Mat}}\def\R{\mathbb R}$Let $A \in \Mat_{n+1,m}(\R)$ given by
$$A = \begin{pmatrix} x_1 & x_2 & \cdots & x_m\\ 1 & 1 & \cdots & 1 \end{pmatrix} $$
and $b \in \R^{n+1}$ by $b = \binom{0_n}1$. Then, by Farkas,
either (a) for some $\lambda\in\R_{\ge 0}^m$, we have $A\lambda = b$, which is equivalent to $\sum_{i=1}^m ...
1
You should know that the dot product is distributive.
$\left<\mathbf{a}, \mathbf{b} + \mathbf{c} \right> = \left<\mathbf{a}, \mathbf{b} \right> + \left<\mathbf{a}, \mathbf{c} \right>$
So you should be able to obtain two dot products from this expression. Each of them would be what is called "scalar triple products."
How is the result of ...
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