New answers tagged

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I think that the problem is ill-posed. In the sequel, we assume that the $(s_i)$ are non-zero real numbers. Let $A$ be the considered matrix and $B$ be the matrix deduced from $A$ by removing its first row. If there are $i\not=j$ s.t. $s_i=s_j$, then $rank(A)<6$. If the $(s_j)$ are distinct, then is $rank(A)=6$ true ? I think that the answer is yes. It ...


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$R = $ $\begin{pmatrix} a_{11} & a_{21} & ... a_{n1} \\a_{21} & a_{22} & ... a_{n2} \\ a_{n1} & a_{n2} & ... a_{nn} \end{pmatrix}$ $L$ is the lower triangular matrix, $\begin{pmatrix} \ell_{11} & 0 & ...0_n \\ \ell_{21} & \ell_{22} & ...0_n \\ \ell_{31} & \ell_{32} & ...\ell_{3n} \end{pmatrix}$ $L^T$ is the ...


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For the spectral norm, you can write a direct relation between the norm of $B$ and the norm of $A$. Since $A$ is symmetric and positive-definite, the spectral norm of $A$ is just the maximal eigenvalue of $A$. Your $B$ is also symmetric and positive-definite and so its norm also equals to the maximal eigenvalue which will be $||A||^{\frac{1}{n}}$.


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I don't know if there are any nice (i.e. not-too-strong) conditions for the inequality to hold, but I'm sure that it doesn't always hold, even when $C$ is positive definite. Counterexample: \begin{align} A&=A^2=I,\\ B&=B^2=\frac12\pmatrix{1&1\\ 1&1},\\ C&=\operatorname{diag}(1,4). \end{align} In this case, we have $A^2\ge B^2\ge 0$ but ...


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The matrix you wrote down always has two eigenvalues, $2$ and $1$.


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I'd write your labor information matrix as $$L=\left[\matrix{3&1\cr 5&1.5\cr}\right]\ ,$$ so that the first row refers to good$_1$ and the second row to good$_2$. Similarly, you have a plant$_1$ and a plant$_2$ whose labor costs are entered into the first and second columns of $$H=\left[\matrix{22&19\cr18&16\cr}\right]\ .$$ If you now compute ...


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Talking all the time of square matrices of the same order: remember that a matrix $\;Z\;$ is the inverse of a matrix $\;X\;$ if it fulfills $\;XZ=I=$ the unit (or identity) matrix, or equivalently $\;ZX=I\;$. To prove your claim you must show that either $\;(AB)(B^{-1}A^{-1})=I\;$ or $\;(B^{-1}A^{-1})AB=I\;$ . Well, try to do this (remember: matrix ...


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I assume you mean uniqueness up to similarity transformations (i.e. change of basis). Suppose that the linear transformation $T$ is represented by the matrix $\left(\begin{array}{cc}1 & 0 \\ 0 & 2 \end{array} \right)$ with respect to basis $\{e_1,e_2\}$. In other words, $T$ maps any vector $(a,b)=ae_1+be_2$ to $(a,2b)=ae_1+2be_2$. Then, with ...


2

Assume A is some diagnalizable matrix. Then, we can write A = P D$\ P^{-1}$. But, then, we can change the order of our eigenvalues along the diagnal in our matrix D, to produce some other matrix G. But, this corresponds to a change in the order of the eigenvectors in P, which again produces another matrix Q. So, in conclusion , we have A = Q G$\ Q^{-1}$. So, ...


3

The diagonal matrix is unique up to a permutation of the entries (assuming we use a similarity transformation to diagonalize). If we diagonalize a matrix $M = U\Lambda U^{-1}$, the $\Lambda$ are the eigenvalues of $M$, but they can appear in any order.


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I am interpreting the matrix $A$ to be a sequence of matrices, whose limit is taken entrywise, and $B$ likewise. Consider $A_j=\left(\begin{smallmatrix}j&0\\0&1\end{smallmatrix}\right)$. We have $det(A_j)=j$, which approaches infinity as $j\to\infty$. However, $A_j^{-1}=\left(\begin{smallmatrix}\frac{1}{j}&0\\0&1\end{smallmatrix}\right)$, ...


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You want to multiply 3 with 19 and 1 with 16, and then sum it up. A convenient way to express it is $$\begin{bmatrix} 3 & 1\end{bmatrix}\begin{bmatrix} 19 \\ 16 \end{bmatrix}$$


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As one can easily see the property you quoted can fail over a field of positive characteristic. So something similar should be avoided for commutative rings. Let $R$ be a commutative ring with $\operatorname{char}R=0$, and $A\in M_n(R)$ such that $\operatorname{tr}A^k=0$ for $k=1,\dots,n$. Then $A$ is nilpotent. The coefficients $c_i$ of the ...


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Note: I realized eigenplan is the space formed by the eigenvectors for a repeated eigenvalue . So the title you ask a bout the existence of a eigenplan, but in the text you speak a bout invariant hyperplane . Then an invariant hyperplane is not necessarily a eigenplan, the converse is yes. The answer to the question given in the text, is already ...


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That band matrix is almost a Toeplitz matrix. It is also symmetric. Hence, r = [1+4*lambda1, -lambda1, 0, -lambda1, zeros(1,n-4)]; A = toeplitz(r); Update now the northwest and southeast corners: A(1,1) = 1+2*lambda1; A(n,n) = 1+2*lambda1;


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Another way to do it is by outer product: > a = [1 2; 3 4; 5 6; 7 8] > b = [1 -1] > a - ones(4,1)*b This easily gets nastier to express with outer products if you start having 3,4 or even more dimensions and want to replicate any subset of them though.


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Removing a dominated strategy does not mean that the strategy goes away. It just means that it would be very irrational for player $i$ to play the dominated strategy because if you're Player I, then you would receive a lower payoff, and if you're Player II, you would get hurt more. (Since in zero-sum games, Player I's gain is Player II's loss.) So you ...


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You can write it as a sum of "diag" commands. For example the code diag([1,1,1],2) + diag([2,2,2,2],1) + diag([3,3,3,3,3],0) gives $$\left[\begin{array}{ccccc}3&2&1&0&0\\0&3&2&1&0\\0&0&3&2&1\\0&0&0&3&2\\0&0&0&0&3\end{array}\right]$$ of course you can just replace the vectors ...


1

Your characteristic polynomial should be of the form $w^2-2\cos\theta \; w + 1$. Solve it to get your eigenvalues, $w$.


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Actually the two eigenvalues of your rotation matrix are complex (and not real), and more specifically they are $e^{i\phi}$ and $e^{-i\phi}$. Try to compute again (and carefully) the characteristic polynomial to find this result.


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Note that $(\cos\phi-a)^2 \geq 0$ and $\sin^2\phi\geq 0$. Therefore the equation $(\cos\phi-a)^2+\sin^2\phi=0$ holds if and only if $\cos\phi=a$ and $\sin\phi=0$. In particular, $\phi$ must be an integer multiple of $\pi$, and in that case the eigenvalue of $A$ will be $\cos\phi$ (which will be $1$ or $-1$ depending on the value of $\phi$). Otherwise, there ...


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How did you get $a=2\cos \phi$? Try computing the determinant again, carefully, and then solving for all possible $a$. You are right that since $1 = \det I = \det(A^TA) = (\det A)^2$, and since $A$ is real, the product of the eigenvalues must be 1 or -1. But this does not mean that the eigenvalues themselves must be 1 or -1...


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For any $n \times n$ (complex) matrix $A$, there are always invariant subspaces of $\mathbb C^n$ of all dimensions $\le n$. If $A$ is upper triangular, the $k$-dimensional subspace spanned by the first $k$ standard unit vectors is invariant. And every square matrix is similar to an upper triangular matrix.


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If $Av=\lambda v$ and $Aw=\lambda w$, then for any linear combination $\alpha v+\beta w$ we have $$ A(\alpha v+\beta w)=\alpha Av+\beta Aw=\alpha\lambda v+\beta\lambda w=\lambda(\alpha v+\beta w). $$ In words, a linear combination of eigenvectors for the same eigenvalue is again an eigenvector for that eigenvalue. That said, it could happen that no such ...


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A multilinear map is a function which is linear in all arguments, i.e. of the form (for three vector variables in two dimensions) $$m(\mathbf p,\mathbf q,\mathbf r)=\\Ap_xq_xr_x+Bp_yq_xr_x+Cp_xq_yr_x+Dp_yq_yr_x+Ep_xq_xr_y+Fp_yq_xr_y+Gp_xq_yr_y+Hp_yq_yr_y.$$ It is possible to perform linear regression on such a model, but this is uncommon.


3

well, a diagonal matrix with nonzero entries along the main diagonal is always invertible, so to answer the original question there is no rule of thumb. Furthermore, a matrix that is all nonzero may be non-invertible. You would need to look at the rank of the matrix (or many, many other things, like the determinant)


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Assuming you mean Multivariate Linear Regression by multi-linear coefficient: In Multiple Linear Regression you have multiple predictors /independent variables ($x_1,\cdots,x_n$) and only one depentend variable $y$: $$y=\beta_0+\beta_1x_1+\cdots+\beta_nx_n$$ In Multivariate Linear Regression you have multiple predictors ($x_1,\cdots,x_n$) and multiple ...


1

For $\lambda=-3$ we consider $A-\lambda I = A-(-3)I=A+3I$ $\begin{bmatrix}5&8&16\\ 4&1&8\\ -4&-4&-11\end{bmatrix} + \begin{bmatrix}3&0&0\\0&3&0\\0&0&3\end{bmatrix} = \begin{bmatrix} 8&8&16\\4&4&8\\-4&-4&-8\end{bmatrix}$ We try to ask: what is the eigenspace for the eigenvalue $-3$? ...


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Unless I'm mistaken, the matrix for the eigenvalue -3 should have rows 8,8,16; 4,4,8; and -4,-4,-8. The second and third rows are multiples of the first. Appropriate row operations eliminate them. You end with the equation x+y+2z=0, which has solutions equal to the span of the vectors (-1,1,0) and (-2,0,1). I hope this helps.


2

Note that the eigenvalues of $P$ are real and in $[-3,3]$. Let $U_n$ be the matrix that is derived from $A_n$ by putting the entries $[1,n],[n,1]$ equal to $0$. Then $\det(A_n\pm B_n-\lambda I_n)=\det(A_n-\lambda I_n)\pm \det(U_{n-1}-\lambda I_{n-1})=p_n(\lambda)\pm q_{n-1}(\lambda)$. The roots of $p_n$ are $2\cos(\frac{2\pi j}{n}),j=1,\cdots,n$. The ...


2

By computing $Q(-e_i)$ for all $i$ you can see that all of $Q$'s entries must be nonpositive. This means that two columns of $Q$ are orthogonal only if they are structurally orthogonal; since all columns must have at least one nonzero entry, this implies that all of the columns have exactly one nonzero entry. Therefore the only $Q$ satisifying your ...


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Let $A,B,C,D,E$ be matrices. If $A = BCDE$, then $E=D^{-1}C^{-1}B^{-1}A$, (notice the order of the matrices) assuming $B$, $C$, and $D$ are invertible. If they're not invertible you probably can't get $E$ out as multiplication by noninvertible matrices is generally a lossy process (to borrow a data compression term).


2

Diagonalization is shearing any source of origin the x directional and y directional matric elements of rows and column from a source of for example radiation of light rays into squeezing along hypotenuse out of a right angled triangle sides as such resultant a+ib module of shearing.This happens as a result of stretching.It really bends a light ray by ...


5

The ranks you have been given allow you to recover the geometric multiplicities of the eigenvalues $3,1$. There is also an eigenvalue $0$ with geometric multiplicity at least $1$. Work out the sum of the geometric multiplicities of the eigenvalues. Recall algebraic multiplicity is no less than geometric multiplicity. Recall the sum of algebraic ...


2

No this isn't true. Recall the definition: Let $A^* = \bar{A}^\mathrm{T}$ denote the conjugate matrix of $A$. Then $A \in \mathbb{C}^{n,n}$ is called normal if $AA^* = A^*A$. We can easily see that the matrix (taken from here) $A = \pmatrix{ 2&i\\i&-2}$ is orthogonal because $$AA^\mathrm{T} = \pmatrix{ 3&0\\0&3} =A^\mathrm{T}A,$$ but ...


0

Your idea: "Consider sequences $(a_1,a_2,\ldots,a_{2n})$ where $a_i=1$ if $i$ goes into the first row, and $a_i=-1$ if $i$ goes into the second row" is fine. Given such a sequence you can immediately jot down the matrix by writing the $i$s with $a_i=1$ in increasing order in the first row, and then write the remaining $i$s in increasing order in the second ...


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You have defined a map, let's try to show that the map defines a bijection between your set of matrices and the set of sequences of $n$ $1$'s and $n$ $-1$'s such that the partial sums are always non-negative (whatever basic definition of Catalan numbers you have, and there are many, this set is likely easily seen to have size $c_n$, e.g., it is easy to pair ...


0

It is not true that the matrices must be simultaneously diagonalizable. For example, if $M = I$ then all orthogonal matrices commute with $M$ but not all orthogonal matrices are diagonalizable over $\mathbb{R}$. If the distinct eigenvalues of $M$ are $\lambda_1, \dots, \lambda_k$ with multiplicity $n_1, \dots, n_k$, then with respect to an orthonormal basis ...


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The easiest way I know would be to show that $A^2$ and $B^2$ are nonzero, but $A^3=B^3=0$. There is only one $3$-by-$3$ nilpotent matrix of nilpotency index $3$, up to conjugation.


0

Let $J$ be the Jordan canonical matrix that is similar to both A and B. So we have $A=S^{-1}JS$ and $B=R^{-1}JR$ $\Rightarrow J=RBR^{-1}$ $\Rightarrow A=S^{-1}RBR^{-1}S=P^{-1}BP$ This would be a general move you could make. In case you would like to check your work, your $P$ should be \begin{bmatrix} 1 &0&0 \\ ...


1

Recall that the matrix $P$ that you seek is essentially a change-of-basis matrix and that the columns of a transformation matrix are the images of the basis vectors. In this problem, $B$ has a particularly simple form, so let’s start there. The columns of $B$ tell us that, in some basis $\{v_1,v_2,v_3\}$, $v_1$ spans the kernel of the transformation, so ...


0

Someone posted this, but later deleted their post, but I thought it was clever, and probably the closest to what I was looking for, so I thought I'd reproduce it here for reference: Let $A \star B = \pmatrix{A & -I \\ B & I}$, then by the block determinant formula $det(A \star B) = det(A)det(I) - det(-I)det(B) = det(A) + det (B)$.


2

We see that this represents the matrix $$\left[\begin{array}{cccc} k&1&1&1 \\ 1&k&1&1 \\ 1&1&k&1 \\ 1&1&1&k \end{array}\right]\left[\begin{array}{c} a\\b\\c\\d\end{array}\right] = \left[\begin{array}{c} 1\\1\\1\\1\end{array}\right].$$ Hence, we should solve this system of linear equations using the augmented ...


1

We will work in $\mathbb{R}^d$. We will denote $$w^T A w = \left\langle w, Aw \right\rangle,$$where $\left\langle \cdot , \cdot \right\rangle$ is the dot product and for $r \in \mathbb{R}$, $$S(A,r) = \left\{ x \in \mathbb{R}^d \: \middle| \: \left\langle w, Aw \right\rangle = r \right\}.$$ Now, what $S(A,r)$ is like, depends on the choice of $A$ and $r$. We ...


0

As you pointed out, $\det(A) \neq 0$, and then $A^{-1}$ exists. Now: $$A^\top M A=M \Rightarrow \\(A^\top)^{-1}A^\top M AA^{-1}=(A^\top)^{-1}MA^{-1} \Rightarrow\\ M = (A^\top)^{-1}MA^{-1} \Rightarrow\\ M = (A^{-1})^\top MA^{-1}. $$ This suggests that all matrices such that $A = A^{-1}$ and $\det(A) = \pm 1$ satisfy your requirement.


4

Note that since this holds for all positive definite $M$, it holds for $M = I$. So, we must have $$ A^TA = I $$ which is to say that $A$ is an orthogonal (unitary) matrix. We may therefore rewrite the equation as $$ A^{-1}MA = M \implies MA = AM $$ That is, $A$ must commute with all positive definite matrices. However, this means that $A$ must also ...


0

Thanks for the hint, Erick. I now came up with the following argument. Suppose $A$ is a real positive definite symmetric Matrix and that $B$ is a real symmetric logarithm of $A$, i.e. $B$ is real and symmetric and $\exp(B) = A$. We wish to show that $B$ is uniquely determined by $A$. Since $B$ is diagonalizable, it suffices to show that $\sigma(B)$, the ...


1

Let us assume that $B$ is invertible. Write $$\begin{array}{rl} M &= \begin{bmatrix} A + B & A & \ldots & A\\ A & A + B & \ldots & A\\ \vdots & \vdots & \ddots & \vdots\\A & A & \ldots & A + B \end{bmatrix}\\\\ &= \begin{bmatrix} B & O_d & \ldots & O_d\\ O_d & B & \ldots & O_d\\ ...


4

Here's one way to find the eigenvalues: we observe that $A - I$ is a symmetric rank $2$ matrix, so it has exactly $2$ non-zero eigenvalues $\lambda_1,\lambda_2$, and it has $0$ as an eigenvalue exactly $n-1$ times. We note that $$ \lambda_1 + \lambda_2 = \operatorname{trace}(A-I) = n-1\\ \lambda_1^2 + \lambda_2^2 = \operatorname{trace}((A-I)^2) = (n-1)^2 + ...


1

Since $\lambda \,Av_A=X^TXv_A$, multiplying by $X $ on the left you get $$\lambda \,Xv_A=XX^TXv_A=BXv_A. $$ So $Xv_A $ is an eigenvector of $B $ for the eigenvalue $\lambda $. If the multiplicity of $\lambda $ is one, then $v_B $ is a multiple of $Xv_A $.



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