Is this Gram-Scmidt (or an application of) it?

Question

I am given a $2\times 2$ matrix $$\left[ \begin{array}{ccc} a & 0 \\ 0 & b \\ \end{array} \right] $$ where $a,b \in \mathbb{R}$. I was told than an orthnormal basis for the colums of this matrix is $\frac{1}{\sqrt{a}}e_1$ and $\frac{1}{\sqrt{a}} e_2$. How does one show this? At first I thought it was an application of Gram-Schmidt, but I am not sure how to apply it.

Answer 1

If you meant an orthonormal basis for the columns space, and assuming $\;a,b\neq0\;$, then it could be

$$\left\{\;\binom10\;,\;\;\binom 01\;\right\}$$

Of course, the above is not the only orthonormal basis for that space.

Answer 2

There is something incorrect in the statement. I'll first explain why.

In order for a basis to be orthonormal, the basis vectors need to be of length one in addition to being orthogonal.

The notation $e_1$ usually means a vector of all 0's and 1 in the first spot. In general $e_i$ is a vector of all 0's with a 1 in the $i^{th}$ spot.

Therefore the statement that $\frac{1}{\sqrt{a}} e_1 $ and $ \frac{1}{\sqrt{a}} e_2$ are orthnormal is incorrect because each of these vectors have length $$\|\frac{1}{\sqrt{a}} e_1\|_2= \sqrt{{\frac{1}{\sqrt{a}}}^2+0^2+...+0^2}=\sqrt{\frac{1}{a} }=\frac{1}{\sqrt{a}}$$

which is only 1 when $a=1$. $ \frac{1}{\sqrt{a}} e_2$ is similar.

Since a basis is closed under multiplication, any multiple of a basis element can be treated as the same basis element. For example, $$ \left(\begin{array}{c} a\\ 0 \end{array}\right) \text { and } \left(\begin{array}{c} 1\\ 0 \end{array}\right) $$ can be consider the same because $$ \left(\begin{array}{c} a\\ 0 \end{array}\right) = a \left(\begin{array}{c} 1\\ 0 \end{array}\right). $$

Looking at $$\left[\begin{array}{cc} a & 0\\ 0 & b \end{array}\right],$$ $$ \left(\left\{ \begin{array}{c} a\\ 0 \end{array}\right\} ,\left\{ \begin{array}{c} 0\\ b \end{array}\right\} \right) $$ Is an orthogonal basis, but it isn't an orthonormal one. In order to make it orthonormal you need to divide each vector by its length. (In this case, their lengths are just $a$ and $b$.) Thus your orthonornmal basis is $$ \left(\left\{ \begin{array}{c} 1\\ 0 \end{array}\right\} ,\left\{ \begin{array}{c} 0\\ 1 \end{array}\right\} \right) .$$

Gram-Schmidt could be used here to get an orthonormal basis. However, since this example is relatively simple, we don't need such a complicated process. Instead, we can get the orthonormal basis using the thought process I described.