If every non-zero vectors be the eigenvector of a real matrix $A$, prove that $A$ is the scalar matrix $\lambda I_n$.

Question

If every non-zero vectors in $\mathbb{R}^n$ be the eigenvector of a real $n \times n$ matrix $A$ corresponding to a real eigenvalue $\lambda$, prove that $A$ is the scalar matrix $\lambda I_n$.

I have no idea to start with. Please help me to solve it.

Hint: The matrices $A$ and $\lambda I$ are equal and only if they represent the same linear transformation. — hmakholm left over Monica, May 22 '16 at 13:46

Emilio Novati · Accepted Answer · 2016-05-22T14:52:00.290

5

Hint:

$Ax=\lambda x \quad \forall x \in \mathbb{R}^n \quad \iff \quad (A-\lambda I)x=0 \quad \forall x \in \mathbb{R}^n$

edited May 22 '16 at 14:52

answered May 22 '16 at 13:50

Emilio Novati

62,675

1

Ok. Its the definition. What will be the next step... – user1942348 May 22 '16 at 13:52
The second line is not simply a matter of definition. The next step is to say if $B=A-\lambda I$ is a matrix such that $Bx=0$ for every $x$, then... – Ben Grossmann May 22 '16 at 14:08
Then its a homogeneous system. It has zero solutions and has non-zero solutions iff $rank(B)<n$. What you actually want to conclude, please. – user1942348 May 22 '16 at 14:48
Please explore the hints a bit more. – user1942348 May 22 '16 at 15:17
The hint is that $(A-\lambda I)x=0$ is not a system ( to be solved for some $x$) but an identity that it's true for all vectors. – Emilio Novati May 22 '16 at 15:33
@EmilioNovati I am in confusion why $(A-\lambda I)x=0$ is an identity? – user1942348 May 22 '16 at 15:42
It say that $(A-\lambda I)x=0$ for all $x$, and this is possible only if $(A-\lambda I)$ is the null matrix. – Emilio Novati May 22 '16 at 15:46
How can you prove the stronger statement: If every nonzero vector in $\mathbb{R}^n$ is an eigenvector (but I am not telling you anything about the number of eigenvalues) of $A$, then there exists some scalar $\lambda$ such that $A=\lambda I$. – Jeppe Stig Nielsen Feb 25 '21 at 09:53

score 2 · Answer 2 · answered May 22 '16 at 13:59

Let $x$ be a non zero vector, $A(x)=\lambda x$, If $y=hx$, $A(y)=A(hx)=hA(x)=h\lambda x=\lambda y$, if $x,y$ are linearly independent, $A(y)=dy$, $A(x+y)=e(x+y)=A(x)+A(y)=\lambda x+dy$. This implies that $(e-\lambda)x+(e-d)y=0$ since $x,y$ are linearly independent, $e-\lambda=0$ and $e-d=0$, thus $e=\lambda=d$. done.

If every non-zero vectors be the eigenvector of a real matrix $A$, prove that $A$ is the scalar matrix $\lambda I_n$.

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