Tag Info

Hot answers tagged

14

Suppose that $A=I+3B$ is torsion. Then, note that it's eigenvalues must all be of unit length. But, the eigenvalues of $I+3B$ are precisely those of the form $1+3\lambda$ for $\lambda$ an eigenvalue of $B$. So, in particular, we see that for any eigenvalue $\lambda$ of $B$ that $$|3\lambda|=|1+3\lambda-1|\leqslant |1+3\lambda|+1=2$$ So, ...


9

How many quadruples $(a,b,c,d)$ are there in which $ad=bc$? Case 1, $a=0$. Then $d$ is arbitrary and $b$ or $c$, or both, must be zero. There are $p$ choices for $d$ and $2p-1$ for $b$ and $c$. Case 2, $a\ne0$. Then $b,c$ are arbitrary and $d=bc/a$. There are $p-1$ choices for $a$ and $p^2$ for $b$ and $c$. Total, $$p(2p-1)+(p-1)p^2=p^3+p^2-p\ ,$$ and ...


4

Nope, the statement is not true in this form. A simple example is: $$ B=\begin{bmatrix}1\\1\end{bmatrix},\quad A=\begin{bmatrix}1&0\\0&0\end{bmatrix}. $$ Then $\mathrm{rank}(B)=1$ and $B^TAB=1$ is non-singular, but $A$ is singular. If $$ B=\begin{bmatrix}1\\1\end{bmatrix}, \quad A=\begin{bmatrix}1&0\\0&-1\end{bmatrix}, $$ then ...


3

There is such a thing, at least over the reals. Suppose $m>n$. Then an $m\times n$ matrix has full rank if and only if it contains an $n\times n$ submatrix of full rank. Let $A$ be an $m\times n$ matrix and let $A_1,\dots,A_N$ be its $n\times n$ submatrices. (The exact value of the number $N$ is irrelevant here; it only depends on $m$ and $n$.) Now let ...


3

$cA_{n\times n}^0 = cI_{n\times n}$, where $c$ is the constant term. In your case, $c=1$, so we add the $2\times 2$ identity matrix to replace $1$. Your final function evaluated at $A$ is given by $$p(A) = 2A^2 - A + I_{2\times 2}$$ Can you take it from here?


3

Since $A^3=A^2$ then the polynomial $P=x^3-x^2=x^2(x-1)$ annihilates $A$ and since $A$ isn't diagonalisable then the polynomial $x(x-1)$ with simple roots doesn't annihilate $A$ so $P$ is the minimal polynomial of $A$ because $A^2\ne0$ and then $A$ is similar to ...


3

It depends on what "easy way" means. There is no known shortcut for determinants of orthogonal matrices, but most known algorithms will run faster for them. This is not detected by simple complexity estimates in terms of matrix size only. Such estimates assume that arithmetic operations are performed in constant time regardless of the size of numbers ...


3

Warning: This is not the shortest proof. The simplest way to prove the desired identity is via induction. Instead, I'll give a demonstration of two different techniques: I'll determine $B^N$ by first diagonalizing $B$, and then show via a formal power series that the expression so obtained also my be found from desired expansion given above. Finding ...


2

You used the wrong transformation. Why not use $y=Ax$? In this case, you are looking at: $$ \int_{K'}\|y\|^2d(A^{-1}y)=|A|^{-1}\int_{K'}\|y\|^2dy, K'=\{y\in\mathbb R^n:\|y\|\leq 1\} $$ Basically, you want to compute the integration on a $(n-1)$-ball. So...spherical coordinates will be a perfect choice right? According to Wikipedia, we have: $$ dy=dy_1\cdots ...


2

$$40800 = \det \left( \begin{array}{cccc} 15 & 11 & 5 & 4 \\ 10 & 1 & 14 & 9 \\ 7 & 8 & 3 & 16 \\ 2 & 13 & 12 & 6 \\ \end{array} \right)$$


2

Hint : Let $T$ and $S$ be two linear transformation from $ V \rightarrow V$ with matrices $M_1$ and $M_2$. What will happen to $S \circ T$ if range of $T$ is contained in the Kernal of $S$ ?


2

The first algorithm computing the determinant faster than in $O(n^{3})$ time (his algorithm worked in $O(n^{\ln 7/\ln 2})$ time) was given by Volker Strassen in this classical paper. Therefore, $O(n^{\ln 7/\ln 2})$ time suffices to check whether a given number is an eigenvalue or not. In fact, the problem of computing the determinant has asymptotically the ...


2

So, you can write any matrix in $E$ as $$ M = I + \underline{a}\underline{a}^T - \sum_{k=1}^{n} \underline{e}_k\underline{e}_k^Ta_k^2 $$ where $\underline{a}$ is a vector with components in $[-1,1]$. If you use the $2$-norm it's going to be complicated, I think, cause you will need to solve a min-max problem. $1$-norm and $\infty$-norm don't look much ...


2

NOTE: This answer goes with the original question of "is the null space contained in the column space?", but I think it's still worth leaving it here. It does, however contain the answer to the new version of the question of "is the null space contained in the row space?". The answer is that the null space must be orthogonal to the row space and therefore ...


2

Even though one may say $x_1$ is not contributing anything but this very fact makes it a free variable. So in your case both $x_1$ and $x_4$ are free variables. The remaining are pivot variables. So what you are doing is correct. You can have $x_1=1$ and $x_4=0$ for the second special solution. Note: Another way to think about it is that $Ax$ is also the ...


2

If $P(t)=P$ is constant and $R(0)=I$ then $R(t)=e^{tP}$, more generally, if $P(t)$ commute with each other for different $t$ then $R(t)=e^{\int P(t)dt}$. In both cases different $R(t)$ commute with each other. But in general $R(t)$ is not of this form, it's what is called an ordered exponential. The problem is that $e^{P(t)}$ do not commute with each other ...


2

Sure, with $m$ the smallest eigenvalue. If $A$ is strictly positive definite then $m>0$, and $A-mI$ annuls the $m$ eigenvectors, so it's only semidefinite.


2

Here's a proof that's basis free. It works with $3$ replaced by any odd prime. Let $A$ be a finite order element of the kernel. The first thing to realize is that $A$ has order a power of $3.$ To see this one observes that since $\langle A \rangle$ is finite and $\bigcap \ker(SL_n(\mathbb{Z}) \rightarrow SL_n(\mathbb{Z}/3^d)) = 1,$ it must be the case ...


2

The desired sum of all (leaf) path scores can be regrouped, using the distributive law, into an expression matching the structure of the tree:       (I will write $V_1$ instead of $MWV_1$.) $$ \left[\left[\left[\left[ V_1(0) \right] V_1(1) + V_2(0) \right] V_1(2) + \left[ V_1(0) \right] V_2(1) \right] V_1(3) + \left[\left[ V_1(0) \right] ...


2

Hint: If you are given a proof for those estimates, look at them. The proofs contain some other inequalities that you may be more familiar with. When does equality hold in them? A different kind of hint: What special kind of vectors can you come up with in $\mathbb R^n$? What are some simple nonzero vectors? The first inequality is not optimal and thus ...


2

Let $A\in F(m,n)$. Since $A$ has rank $m$, it has some $\;m\times m$ nonvanishing minor. As the determinant function is continuous, the same minor remains nonzero under small perturbations. Obviously, no element in $F(m,n)$ has rank greater than $m$, and we are done.


2

Pretending $A$ is a constant is not how the proof is done, but it is one way to think about it. The proof is achieved by using the definition of matrix multiplication: Let $a_{ij}$ be the element in the $i$th row and $j$th column of $m \times n$ matrix $A$, and let $\mathbf{v}$ and $\mathbf{w}$ be two column $n$-vectors with coordinates $v_j$ and $w_j$. ...


2

Since you only need a subset for this example- we can "cheat" a bit. We shall define our subset $U$ to be the union of $x$ axis, and the $y$ axis, i.e. $U$= $\{(x,0):x \in \mathbb{R}\} \cup \{(0,y):y \in \mathbb{R}\}$


2

The first problem is calculating $$\begin{bmatrix}40 & 60 & 80 & 80\end{bmatrix} \begin{bmatrix}0.3 \\ 0.4 \\ 0.2 \\ 0.5\end{bmatrix}$$ which is a $1 \times 4$ matrix times a $4 \times 1$ matrix, so the result is a $1 \times 1$ matrix (scalar). The second is calculating $$\begin{bmatrix}a \\ 2a \\ 3a \end{bmatrix} \begin{bmatrix}1 & 0 & ...


2

If $c_1,\ldots c_n$ are the entries on the diagonal of an upper triangular matrix $A$, then the characteristic polynomial is $p(x) = (x-c_1) \cdots (x-c_n)$. $A-c_iI$ is upper triangular with the $i$th diagonal entry being zero. See what happens when you then compute $p(A)=(A-c_1 I) \cdots (A-c_nI)$.


1

Look at the first block, which is of the form $C = [\alpha^2 \ \alpha \beta; \alpha \beta \ \beta^2]$. Let $z = [x; y]$ be a column $2$-vector. Then: $z' C z = [\alpha x; \beta y]' [1 \ 1; 1 \ 1] [\alpha x; \beta y] \geq 0$, because the matrix $[1 \ 1; 1 \ 1]$ is sp(semi)d. The same happens with the $3\times3$ block. Therefore, your matrix is sp(semi)d.


1

If $V$ and $W$ are finitely dimensional $K$-vector spaces of the same dimension and $f:V\times W\to K$ is a bilinear pairing, then the matrix of $f$ in the bases $\{x_1,\dots,x_n\}$, respectively $\{y_1,\dots,y_n\}$ of $V$, respectively $W$ is given by $(f(x_i,y_j))_{1\le i,j\le n}$. In your case $E(r_i\otimes s_j,r_k\otimes s_l)=B(r_i,r_k)C(s_j,s_l)$ and ...


1

Hint: (Rank part) Consider the formula about rank of product of matrices $$\text{rank}(AB)\leq\min\{\text{rank}(A),\text{rank}(B)\}$$ (Eigenvalue part) Note that $(ab^T)x=a(b^Tx)=(b^Tx)a=\lambda x$ implies $a\parallel x$.


1

$A^t = A^{-1} = (\operatorname{cofactor}(A))^t $ implies $A = \operatorname{cofactor}(A)$ and I am not sure why $A$ should have determinant $1$. It seems the conclusion is true for any real orthogonal matrix.



Only top voted, non community-wiki answers of a minimum length are eligible