5
votes
Let $A,B$ be $2$ square matrices such that $A+AB-2B=0$. Prove $2$ isn't an eigenvalue of $A$.
$A+AB-2B=0$
$A+AB-2B-2I=-2I$
$(A-2I)(B+I)=-2I$
$\therefore A-2I$ is non-singular.
$\therefore$ for any $v \neq 0,$
$(A-2I)v \neq 0$
$Av \neq 2v$
$\therefore 2$ cannot be an eigenvalue.
- 2,908
3
votes
Accepted
Let $A,B$ be $2$ square matrices such that $A+AB-2B=0$. Prove $2$ isn't an eigenvalue of $A$.
Recall that since the matrices are square, $A$ and $A^T$ have the same eigenvalues.
In particular for $v\neq 0$ such that $A^Tv=2v$ then
$$A^Tv+B^TA^Tv-2B^Tv=2v+2B^Tv-2B^Tv=2v\neq 0$$
which is a ...
- 27.8k
2
votes
Accepted
Prove with induction if $A= \begin{pmatrix}2&1\\-1&0 \end{pmatrix}$ $\forall n \in \mathbb{N}$ $A^n= \begin{pmatrix} n+1&n\\-n&1-n \end{pmatrix}$
To prove by induction that for the matrix
$$A = \begin{pmatrix}
2 & 1 \\
-1 & 0
\end{pmatrix}$$
for all natural numbers $n$,
$$A^n = \begin{pmatrix}
n + 1 & n \\
-n & 1 - n
\end{...
- 7,104
2
votes
Prove with induction if $A= \begin{pmatrix}2&1\\-1&0 \end{pmatrix}$ $\forall n \in \mathbb{N}$ $A^n= \begin{pmatrix} n+1&n\\-n&1-n \end{pmatrix}$
this question illustrates how finding the Jordan form, including the change of basis matrix, cleans things up.
$$\left(
\begin{array}{rr}
1 & 0 \\
-1 & 1 \\
\end{array}
\right)
\left(
\...
- 137k
1
vote
Accepted
(Linear Independence) Show that for two vectors $v,w \in \mathbb{R}^n $, the conditions (i), (ii), (iii) are equivalent
(i) $\implies (ii)$
if $w\neq 0$ and $v=\rho w$ since $\Bbb R$ is a field exist the inverse of $\rho$ namely $\frac{1}{\rho}$ but this lead to a contradiction since $\frac{1}{\rho} \in \Bbb R, w = \...
- 790
1
vote
Are all nilpotent matrices strictly upper triangularizable (over the complex and real fields)?
This answer would fit better as a comment to Eric Wofsey but it does give a very elementary proof to the original question by showing that if two real matrices are conjugates over complex matrices, ...
- 46
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