# linear algebra proof

• Mar 30th 2010, 01:30 AM
deniselim17
linear algebra proof
I have no idea where to start to prove this.

Let $A=I_{n} -u^{T}u \in M_{n}(R)$ where $I_{n}$ is the identity, $u$ is a non-zero row vector in $R^{n}$ and $u^{T}$ denotes transpose of $u$. If $A$ is symmetric, show that $u^{T}u=I_{n}$.

$I_{n}, u^{T}u$ both symmetric, then of course $A$ is symmetrc. But how to show $u^{T}u$ is equal to identity?
• Mar 30th 2010, 04:40 AM
tonio
Quote:

Originally Posted by deniselim17
I have no idea where to start to prove this.

Let $A=I_{n} -u^{T}u \in M_{n}(R)$ where $I_{n}$ is the identity, $u$ is a non-zero row vector in $R^{n}$ and $u^{T}$ denotes transpose of $u$. If $A$ is symmetric, show that $u^{T}u=I_{n}$.

$I_{n}, u^{T}u$ both symmetric, then of course $A$ is symmetrc. But how to show $u^{T}u$ is equal to identity?

Without any further conditions this is false: take for example $u=(12)$, then:

$A= I-u^tu=\begin{pmatrix}1&0\\0&1\end{pmatrix}-\begin{pmatrix}1&2\\2&4\end{pmatrix}=\begin{pmatri x}0&\!\!\!-2\\\!\!\!-2&\!\!\!-3\end{pmatrix}$ , so $A$ is symmetric but $u^tu$ is not the unit matrix.

As you say: sum/difference and product of symmetric matrices is symmetric, so the above difference will always be a symmetric matrix, no matter what $u^tu$ is.

Tonio
• Mar 30th 2010, 05:47 PM
deniselim17
Quote:

Originally Posted by tonio
Without any further conditions this is false

Did you mean that $u^{T}u=I_{n}$ is false?
• Mar 30th 2010, 06:52 PM
tonio
Quote:

Originally Posted by deniselim17
Did you mean that $u^{T}u=I_{n}$ is false?

Yes.

Tonio