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

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 $\lambda I_n$.

• 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

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