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If $A + tB$ is similar to $A$ for infinitely many values of $t$, where $A$ is diagonalizable, is $B$ necessarily equal to $0$?

For the non-diagonalizable case, take $$ A = \begin{bmatrix} 1 & 1 \\0 & 1 \end{bmatrix}, \qquad B = \begin{bmatrix} 0 & 1 \\0 & 0 \end{bmatrix}. $$ For all $t$ except $t=-1$, we have $...
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1 vote

Does $X, X^t$ have same operator norm.

\begin{align} \left\|X\right\| &= \max_{\left\|h\right\|=1} \left\|Xh\right\|\\ &= \max_{\left\|h\right\|=1} \left\|\overline X^T h\right\|\\ &= \max_{\left\|\overline h\right\|=1} \left\|\...
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  • 5,308
0 votes

Show that $\log |\Sigma|+\text{Tr}(\Sigma^{-1}V)$ is uniquely minimized

Another argument you can use the concavity of $\Sigma \to \log \left|\Sigma\right|$ (see here). Since $$f\left(\Sigma^{-1}\right) = -\log \left|\Sigma\right| + \text{Tr}\left(\Sigma V\right)$$ then $\...
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  • 5,308
1 vote
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When does a real symmetric matrix have $LDL^{T}$ decomposition? And when is the $LDL^{T}$ decomposition unique?

Let us answer the questions one by one: Point1: The factorization $LDU$ relies on the factorization $LU$ and then we use $D$ to make $U$ as an upper triangular matrix with $1$'s in the main diagonal. ...
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0 votes

Decomposition for non symmetric matrix using left and right eigenvectors

For a square matrix $A$, $x$ is a right-eigenvector if $Ax = \lambda x$ for some scalar $\lambda$ and a left-eigenvector if $A^Tx = \lambda x$ for some scalar $\lambda$. $A^T$ denotes the transpose of ...
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2 votes
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How unique is thin/reduced QR decomposition without $R_{ii}>0$ condition?

The sticking point in my proof is the assumption that the 𝑅 matrices are invertible (or simply that they have nonzero diagonals). Notice for any choice of $k$ $0\lt \det\big(A^T A\big) = \det\big(...
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1 vote
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To compute Hermite Normal Form H and U

To undestand how the process works, it is best illustrated with a non-squared matrix. As square matrices are easier to work with, it doesn't show the full extend of the process. Process: The intuitive ...
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  • 116
1 vote

Gram matrix of QR decomposition

The projection $\hat x$ of $x$ onto the image of $A$ is $$ \hat x = A(A^T A)^{-1} A^T x. \tag{1} $$ so $y = x - 2 A(A^T A)^{-1} A^T x $. This is a well known formula in the topic "linear ...
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0 votes
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Simple eigendecomposition of matrix

In general, you can't. We know the eigenvectors of $\bf P D P^{-1}$ are the columns $\bf v_i$ of $\bf P$, $i = 1, \ldots, n$. Let's call the corresponding eigenvalues $\lambda_i$. Now, observe that, ...
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  • 330
0 votes

Cholesky decomposition of the inverse of a matrix

To add to previous answers, if we view $X$ as covariance matrix of data, the relationship between two decompositions reduces to relationship between coefficients of "right-to-left" ...
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1 vote

Show that no $n \times n$ real matrix $A^2$ can be a diagonal matrix with distinct negative entries

Here is a simpler proof that requires only that $a_1$ be different from the other $a_j$'s. Supposing that such a matrix $A$ exists, let $p$ be a polynomial such that $p(-a_1)=1$, and $p(-a_j)=0$, for ...
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2 votes

Show that no $n \times n$ real matrix $A^2$ can be a diagonal matrix with distinct negative entries

$A^2=\begin{bmatrix} -a_1 \\ &\ddots \\ & & -a_n\end{bmatrix}$ implies, working over $\mathbb C$, that each eigenvalue of $A$ satisfies $\lambda_k^2 \lt 0$, i.e. each eigenvalue must be ...
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3 votes

Show that no $n \times n$ real matrix $A^2$ can be a diagonal matrix with distinct negative entries

Let's suppose that $$A^2=\begin{pmatrix} -a_1 \\ &\ddots \\ & & -a_n\end{pmatrix}$$ Then the characteristic polynomial of $A^2$ is $P(X)= (-1)^n\prod_{k=1}^n (X+a_k)$, so by Cayley-...
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3 votes
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Could the product of a skew-symmetric matrix and an invertible matrix be nilpotent?

In fact, for any matrix $A$, skew-symmetric or otherwise, there is some invertible matrix $B$ such that $AB$ is nilpotent if and only if $A=0$ or $A$ is not invertible. First and foremost, it's an ...
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1 vote
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Rotation matrix decomposition into mixed local/global Euler angles

In any 3 angle rotation scheme, there are actually four reference frames at play. Lets define: Frame Description $G$ Global frame before rotations $L_1$ Local frame after 1st rotation $L_2$ Local ...
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  • 86
1 vote
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Cholesky inverse

To solve $PDP^T = LXL^T$ with matrices as specified, compute the diagonal matrix $R$ such that $RR^T = D$ by taking the square roots of the diagonal entries. then you have: $PRR^TP^T = LCC^TL^T$ where ...
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