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

Problems and solutions in Linear Algebra

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.

 
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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*}


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  1. 08/16/2017

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