# The Inner Product on $\R^2$ induced by a Positive Definite Matrix and Gram-Schmidt Orthogonalization

## Problem 539

Consider the $2\times 2$ real matrix
$A=\begin{bmatrix} 1 & 1\\ 1& 3 \end{bmatrix}.$

(a) Prove that the matrix $A$ is positive definite.

(b) Since $A$ is positive definite by part (a), the formula
$\langle \mathbf{x}, \mathbf{y}\rangle:=\mathbf{x}^{\trans} A \mathbf{y}$ for $\mathbf{x}, \mathbf{y} \in \R^2$ defines an inner product on $\R^n$.
Consider $\R^2$ as an inner product space with this inner product.

Prove that the unit vectors
$\mathbf{e}_1=\begin{bmatrix} 1 \\ 0 \end{bmatrix} \text{ and } \mathbf{e}_2=\begin{bmatrix} 0 \\ 1 \end{bmatrix}$ are not orthogonal in the inner product space $\R^2$.

(c) Find an orthogonal basis $\{\mathbf{v}_1, \mathbf{v}_2\}$ of $\R^2$ from the basis $\{\mathbf{e}_1, \mathbf{e}_2\}$ using the Gram-Schmidt orthogonalization process.

## Proof.

### (a) Prove that the matrix $A$ is positive definite.

We prove that for every nonzero vector $\mathbf{x}=\begin{bmatrix} x \\ y \end{bmatrix}\in \R^2$, we have $\mathbf{x}^{\trans} A \mathbf{x} > 0$.
We have
\begin{align*}
\mathbf{x}^{\trans} A \mathbf{x}&=\begin{bmatrix}
x & y
\end{bmatrix} \begin{bmatrix}
1 & 1\\
1& 3
\end{bmatrix}\begin{bmatrix}
x \\
y
\end{bmatrix}=\begin{bmatrix}
x & y
\end{bmatrix}\begin{bmatrix}
x+y \\
x+3y
\end{bmatrix}\6pt] &=x(x+y)+y(x+3y)=x^2+2xy+3y^2\\ &=x^2+2xy+y^2+2y^2=(x+y)^2+2y^2. \end{align*} Since \mathbf{x}\neq \mathbf{0}, at least one of x, y is nonzero. Thus the last expression is always positive. Hence A is a positive definite matrix. ### (b) Prove that \mathbf{e}_1, \mathbf{e}_2 are not orthogonal in the inner product space \R^2. Note that by post “A Symmetric Positive Definite Matrix and An Inner Product on a Vector Space“, the formula \langle \mathbf{x}, \mathbf{y}\rangle defines an inner product on \R^2. Two vectors \mathbf{x} and \mathbf{y} is said to be orthogonal if \langle \mathbf{x}, \mathbf{y}\rangle=0. The vectors \mathbf{e}_1, \mathbf{e}_2 are not orthogonal with this inner product since \begin{align*} \langle \mathbf{e}_1, \mathbf{e}_2\rangle=\begin{bmatrix} 1 & 0 \end{bmatrix}\begin{bmatrix} 1 & 1\\ 1& 3 \end{bmatrix}\begin{bmatrix} 0 \\ 1 \end{bmatrix}=\begin{bmatrix} 1 & 0 \end{bmatrix}\begin{bmatrix} 1 \\ 3 \end{bmatrix}=1\neq 0. \end{align*} ### (c) Find an orthogonal basis using the Gram-Schmidt orthogonalization process. By the Gram-Schmidt orthogonalization process, we have \begin{align*} \mathbf{v}_1&=\mathbf{e}_1\\ \mathbf{v}_2&=\mathbf{e}_2-\frac{\langle \mathbf{v}_1, \mathbf{e}_2 \rangle}{\langle \mathbf{v}_1, \mathbf{v}_1 \rangle}\mathbf{v}_1 =\mathbf{e}_2-\frac{\langle \mathbf{e}_1, \mathbf{e}_2 \rangle}{\langle \mathbf{e}_1, \mathbf{e}_1 \rangle}\mathbf{e}_1. \end{align*} We compute \begin{align*} \langle \mathbf{e}_1, \mathbf{e}_1 \rangle=\begin{bmatrix} 1 & 0 \end{bmatrix}\begin{bmatrix} 1 & 1\\ 1& 3 \end{bmatrix}\begin{bmatrix} 1 \\ 0 \end{bmatrix}=\begin{bmatrix} 1 & 0 \end{bmatrix}\begin{bmatrix} 1 \\ 1 \end{bmatrix}=1. \end{align*} We also have \langle \mathbf{e}_1, \mathbf{e}_2\rangle=1 from part (b). Thus, we have \begin{align*} \mathbf{v}_2=\mathbf{e}_2-\mathbf{e}_1=\begin{bmatrix} -1 \\ 1 \end{bmatrix}. \end{align*} Thus, the Gram-Schmidt orthogonalization process yields the orthogonal basis \[\mathbf{v}_1=\begin{bmatrix} 1 \\ 0 \end{bmatrix}, \mathbf{v}_2=\begin{bmatrix} -1 \\ 1 \end{bmatrix}.

#### Double Check

Let us verify that $\mathbf{v}_1, \mathbf{v}_2$ are orthogonal by computing their inner product directly as follows.
We have
\begin{align*}
\langle \mathbf{v}_1, \mathbf{v}_2\rangle=\mathbf{v}_1^{\trans} A\mathbf{v}_2=\begin{bmatrix}
1 & 0
\end{bmatrix}\begin{bmatrix}
1 & 1\\
1& 3
\end{bmatrix}\begin{bmatrix}
-1 \\
1
\end{bmatrix}=\begin{bmatrix}
1 & 0
\end{bmatrix}\begin{bmatrix}
0 \\
2
\end{bmatrix}=0.
\end{align*}

### 1 Response

1. 08/16/2017

[…] the post ↴ The Inner Product on $R^2$ induced by a Positive Definite Matrix and Gram-Schmidt Orthogonalization for […]

##### A Symmetric Positive Definite Matrix and An Inner Product on a Vector Space

(a) Suppose that $A$ is an $n\times n$ real symmetric positive definite matrix. Prove that $\langle \mathbf{x}, \mathbf{y}\rangle:=\mathbf{x}^{\trans}A\mathbf{y}$ defines an...

Close