3
votes
Accepted
Prove: if $\text{rank}(A)=\text{rank}(A^2)$ then $Ax=0$ and $A^2x=0$ have the same set of solutions.
$$n-\dim(\ker A^2)=\text{rank}~ A^2=\text{rank}~ A=n-\dim(\ker A)$$
we get
$$\dim(\ker A^2)=\dim(\ker A)\tag{1}$$
We also know:
$$x\in\ker A\Rightarrow Ax=0\Rightarrow A^2x=0\Rightarrow x\in \ker A^2$$...
- 15.7k
3
votes
Prove: if $\text{rank}(A)=\text{rank}(A^2)$ then $Ax=0$ and $A^2x=0$ have the same set of solutions.
Use the fact that ${\rm Ker} A \subseteq {\rm Ker} A^2$ which is true in general. (To see why, if we have $x \in {\rm Ker} A$, then $A x = 0 \implies A^2 x = A (A x) = A (0) = 0$, so that $x \in {\rm ...
- 5,137
2
votes
Accepted
Problem on rank and kernel of linear transformation over polynomial vector space
$V$ is a vector space over $\color{red}{\mathbb{C}}$ and $\dim(V) =n+1$
$T\in\mathcal{L}(V) $ defined by $$T(p(x)) =p'(1) $$
Claim:
$\textrm{range}(T) =\{a_0:a_0\in\Bbb{C}\}$
$\dim(\textrm{null}(T)) ...
- 12.5k
2
votes
Proving that A=A(B−I) and B=B(A−I) implies A is diagonalizable.
The assumptions can be rewritten as $AB=2A$ and $BA=2B.$ Therefore $Bx=0$ iff $Ax=0.$ Hence
$\dim \ker A=\dim \ker B.$ The identity $B(A-2I)=0$ implies that $$\dim \ker B\ge \dim{\rm Im} (A-2I)=n-\dim\...
- 20.8k
2
votes
$W$ is a vector space of all real $n\times n$ matrices $X$ such that $AX=0$ where $A$ is a real $n\times n$ matrix of rank $r$. Find dimension of $W$.
Note that since $A$ is of rank $r$, we can write :
$$ A = P \left(\begin{array}{cc}
\mathcal{I}_r & \mathcal{O}_{r, n-r} \\
\mathcal{O}_{n-r, r} & \mathcal{O}_{n-r, n-r}
\end{array}\right) Q $...
- 1,878
2
votes
Accepted
$W$ is a vector space of all real $n\times n$ matrices $X$ such that $AX=0$ where $A$ is a real $n\times n$ matrix of rank $r$. Find dimension of $W$.
We have $\text{rank}(A)+ \text{nullity}(A)=n$.
Then $\text{nullity}(A)=n-r$.
Let $X\in W$ so that $AX=0$.
Let $X=(C_1,C_2,...,C_n)$ where each $C_i$, $i=1,2,\dots ,n$ is a column vectors of $X$.
Since ...
- 84
2
votes
Is the set of matrices with rank at most $2$ closed?
I do not think that your proof is correct. You claim that there is a "universal sequence of row operations" transforming all matrices into echelon form. But it seems to me that we need ...
- 68.7k
1
vote
Proving that A=A(B−I) and B=B(A−I) implies A is diagonalizable.
If $x\in\ker(A-2I)$ then $Ax=2x$ so
$$2x=A(B-I)x=ABx-2x\quad\Rightarrow\quad x=A(\tfrac14Bx)\in{\rm im}(A)\ ,$$
so $\ker(A-2I)\subseteq{\rm im}(A)$. Conversely, your equations imply $A^2=2A$, so if $...
- 80.3k
1
vote
Accepted
Proof of $\operatorname{rank} \left( X’ V X \right) = \operatorname{rank} (X)$
Here is a proof of the result. I assume that in this context, $V$ being "non-ngeative definite" implies that it is also symmetric (or Hermitian if we're dealing with complex matrices).
We ...
- 216k
1
vote
Accepted
Rank of partitioned matrix
To be precise, you need to prove that $\operatorname{col}(X_1) \cap \operatorname{col}(X_2) = 0$, not $\emptyset$. And yes, this is implied by $X_1 a \ne X_2 b$ for $a, b \ne 0$, since every nonzero ...
- 9,287
1
vote
Is the set of matrices with rank at most $2$ closed?
I'm going to address only your posted question, regarding whether there is a problem with your argument.
Yes, there is a big problem.
It is true that any individual row operation is a continuous ...
- 111k
1
vote
When is ${\rm rank}(A+B)={\rm rank}(A)+{\rm rank}(B)$ true?
Note that the rank is equal to the dimension of the column space (and row space).
$rank(A + B) = \dim col(A + B)$
$\leq \dim (col(A) + col(B))$
$\leq \dim col(A) + \dim col(B)$
$= rank(A) + rank(B)$
...
- 31
1
vote
Prove: if $\text{rank}(A)=\text{rank}(A^2)$ then $Ax=0$ and $A^2x=0$ have the same set of solutions.
One can easily check that $$\rm Im(A)=Im(A^2)\Longleftrightarrow Ker(A)=Ker(A^2)$$
Make use of the formula
$$\rm dim(Ker(A))+dim(Im(A))=n$$
- 21
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