4
votes
Accepted
How we can show this matrix is nilpotent?
In general, suppose that $I-XY=k(I-YX)$ for some complex number $k$ that is not a root of unity. (In your case, $k=-2$.) Since $I-XY$ and $I-YX$ have the same spectrum $S$, we have $S=kS$. As $k$ is ...
- 130k
1
vote
Accepted
Every matrix that commutes with $A \in \Bbb C^{n \times n}$ is a polynomial in $A$
Hint $:$ To disprove $(a)$ and $(b)$ just consider the non-diagonalizable nilpotent matrix (of index $2$) $$A = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}.$$
It's easy to check that all the ...
- 1,929
1
vote
Accepted
Eigenvalues of an $n \times n$ symmetric tridiagonal matrix
The $n \times n$ matrix can be rewritten as a sum of
$n-2$ positive semi-definite matrices (consists of a single
$2\times 2$ non-zero block $\left[\begin{smallmatrix}35 & -35\\-35 & 35\end{...
- 120k
1
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Is this non-symmetric matrix positive definite?
Your definition of positive definiteness of $A$ is equivalent to the usual one applied to the symmetric matrix $\frac{1}{2}(A+A^T).$ Clearly your $A$ cannot be positive definite if for instance $\frac{...
- 660
1
vote
Accepted
Polynomial maps of finite-dimensional vector spaces
$\operatorname{Hom}(V,W)$ refers to the $linear$ maps between vector spaces $V,W$. Although linear maps are always polynomial maps, polynomial maps needn't be linear maps.
In the case that $W=k$ is ...
- 3,820
1
vote
Accepted
if $T: V\to V$ and $ dim(KerT)+dim(ImT)=dimV $ can i say that $ KerT⊕ImT=V $
It is false that $\operatorname{Ker} T \oplus \operatorname{Im} T = V$. In fact, if $V$ is of even dimension, you can even have $\operatorname{Ker} T = \operatorname{Im} T$, by choosing a basis $(e_1, ...
- 1,337
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Projection matrix product under operator norm
The answer is no. As a simple counterexample, consider
$$
A = \pmatrix{1&-1\\0&0}, \quad P = \pmatrix{1&0\\0&0}, \quad B = \pmatrix{1&0\\1&0}.
$$
We find that $APB = P$ and $...
- 214k
1
vote
Accepted
Find the linear transformation in terms of the canonical base
There are two ways of doing this:
As others suggested in the comments --- which given the amount of structure this problem admits is probably the fastest --- one expresses each element of the new ...
- 3,564
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Computing $(A\otimes I + I \otimes A)^{-1} \text{vec}B$
$
\def\LR#1{\left(#1\right)}
\def\op#1{\operatorname{#1}}
\def\vcc#1{\op{vec}\LR{#1}}
\def\trace#1{\op{Tr}\LR{#1}}
\def\qiq{\quad\implies\quad}
\def\p{\partial} \def\grad#1#2{\frac{\p #1}{\p #2}}
\def\...
- 31.4k
1
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Why is span defined as the linear combination with *finite* terms?
One thing to keep in mind: the axioms of a vector space provide a binary sum operation, and so you get linear combinations of the form $$a_1 v_1 + a_2 v_2 = \sum_{i=1}^2 a_i v_i
$$
By applying the ...
- 109k
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