3
votes
Accepted
When do affine transformations preserve ratios of n-dimensional hypervolume?
Yes, and this property follows immediately from applying the multivariable Chain Rule to a general affine transformation. Under the usual identifications, an affine transformation $F : {\bf x} \mapsto ...
- 89.6k
3
votes
Accepted
Prove two functions on an inner product space $S,T: V \to V$ are linear
This is just a slightly expanded version of the comment I made above. You can make real progress on this question just by lining up what is known. We want to show that $T$ is a linear map, that is$$\...
- 549
2
votes
Accepted
Can I use a function argument to prove invertibility of matrices?
Matrices and corresponding linear transformations have a lot of properties in common (I personally like to think of these as fundamentally properties of linear transformations, and that matrices ...
- 194k
1
vote
Try to find all linear transformation satisfy $\mathscr{A}(AB) = \mathscr{A}(B)\mathscr{A}(A)$
Here's another approach:
Suppose $\mathscr{A}$ satisfies the condition and let $\mathscr{B}=\mathscr{A}\circ(-)^T$. Then
$$\mathscr{B}(AB)=\mathscr{B}(A)\,\mathscr{B}(B)$$
so $\mathscr{B}$ is an ...
- 2,282
1
vote
Accepted
A linear map from $\mathbb {R^2}$ to $\mathbb {R^2}$ maps unit circle to unit circle then what can you say about the linear map?
You have proven that $T$ must be an isometry. In particular, $T$ is invertible and its inverse is its adjoint. Indeed, let $A$ the matrix representation of $T$, we must have:
$$(x,y)=(Ax,Ay)=(A^TAx,y)$...
- 8,061
1
vote
Prove that there are no bases in $\mathbb R^3$ in which $f$ has the matrix $B$
If $B$ and $B'$ are bases of $\Bbb R^3$, then what you did shows that the matrix $[f]_B^B$ of $f$ with respect to the basis $B$ cannot possible be equal to $[f]_{B'}^{B'}$. But the idea is to prove ...
- 416k
1
vote
Prove that there are no bases in $\mathbb R^3$ in which $f$ has the matrix $B$
Suppose and $B_1=(v_1,v_2,v_3)$ and $C_1=(w_1,w_2,w_3)$ are ordered basis of $\mathbb{R}^3$ in which $f$ has matrix representation
$$A=\begin{bmatrix}
1&0&2\\
7&2&9\\
1&0&5
...
- 33.6k
1
vote
Accepted
Inductive proof that any nilpotent operator is upper-triangularisable
The idea is nice, but there is a big hidden induction problem.
I translate what you tried to mean in more familiar words to me. Let me know if this was the same your idea.
If $T=0$, then it is clear. ...
- 8,383
1
vote
Prove $ST = 0 \implies ST^* = 0$.
Your proof is correct, and this statement is not true in general for non-normal operators. Let $\{e_0, e_1, \dots\}$ be a basis for your Hilbert space and consider the shift operator $T(e_i) = e_{i+1}$...
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