Let $x$ be a variable and consider the length of the vector $\mathbf{a}-x\mathbf{b}$ as follows.
We have
\begin{align*}
0& \leq \|\mathbf{a}-x\mathbf{b}\|^2=(\mathbf{a}-x\mathbf{b})\cdot (\mathbf{a}-x\mathbf{b})\\
&=\mathbf{a}\cdot \mathbf{a}-\mathbf{a}\cdot x\mathbf{b}-x\mathbf{a}\cdot \mathbf{b}+x^2\mathbf{b}\cdot \mathbf{b}\\
&=\|\mathbf{b}\|^2 x^2-2\mathbf{a}\cdot \mathbf{b}x+\|\mathbf{a}\|^2. \tag{*}
\end{align*}
Note that the last expression is an equation of a parabola (quadratic equation).
Since the parabola is always non-negative, its discriminant $D$ must be non-positive.
Hence we have
\[D=(2\mathbf{a}\cdot \mathbf{b})^2-4\|\mathbf{a}\|^2 \|\mathbf{b}\|^2 \leq 0.\]
It follows that we have
\[(\mathbf{a}\cdot \mathbf{b})^2 \leq \|\mathbf{a}\|^2 \|\mathbf{b}\|^2 .\]
Taking the square root, we obtain the Cauchy-Schwarz inequality
\[|\mathbf{a}\cdot \mathbf{b}|\leq \|\mathbf{a}\|\,\|\mathbf{b}\|.\]
Proof 2
The second proof starts with the same argument as the first proof.
As in Proof 1 (*), we obtain
\[0\leq \|\mathbf{b}\|^2 x^2-2\mathbf{a}\cdot \mathbf{b}x+\|\mathbf{a}\|^2.\]
Now we take
\[x=\frac{\mathbf{a}\cdot \mathbf{b}}{\|\mathbf{b}\|^2}.\]
Then we have
\begin{align*}
0 &\leq \|\mathbf{b}\|^2 \left(\,\frac{\mathbf{a}\cdot \mathbf{b}}{\|\mathbf{b}\|^2} \,\right)^2-2\mathbf{a}\cdot \mathbf{b}\left(\, \frac{\mathbf{a}\cdot \mathbf{b}}{\|\mathbf{b}\|^2} \,\right)+\|\mathbf{a}\|^2\\[6pt]
&=\frac{(\mathbf{a}\cdot \mathbf{b})^2}{\|\mathbf{b}\|^2}-2\frac{(\mathbf{a}\cdot \mathbf{b})^2}{\|\mathbf{b}\|^2}+\|\mathbf{a}\|^2\\
&=-\frac{(\mathbf{a}\cdot \mathbf{b})^2}{\|\mathbf{b}\|^2}+\|\mathbf{a}\|^2.
\end{align*}
It follows that we have
\[(\mathbf{a}\cdot \mathbf{b})^2\leq \|\mathbf{a}\|^2 \|\mathbf{b}\|^2.\]
The Cauchy-Schwarz inequality is obtained by taking the square root as in Proof 1.
Find the Distance Between Two Vectors if the Lengths and the Dot Product are Given
Let $\mathbf{a}$ and $\mathbf{b}$ be vectors in $\R^n$ such that their length are
\[\|\mathbf{a}\|=\|\mathbf{b}\|=1\]
and the inner product
\[\mathbf{a}\cdot \mathbf{b}=\mathbf{a}^{\trans}\mathbf{b}=-\frac{1}{2}.\]
Then determine the length $\|\mathbf{a}-\mathbf{b}\|$.
(Note […]
Inner Products, Lengths, and Distances of 3-Dimensional Real Vectors
For this problem, use the real vectors
\[ \mathbf{v}_1 = \begin{bmatrix} -1 \\ 0 \\ 2 \end{bmatrix} , \mathbf{v}_2 = \begin{bmatrix} 0 \\ 2 \\ -3 \end{bmatrix} , \mathbf{v}_3 = \begin{bmatrix} 2 \\ 2 \\ 3 \end{bmatrix} . \]
Suppose that $\mathbf{v}_4$ is another vector which is […]
Find the Inverse Matrix of a Matrix With Fractions
Find the inverse matrix of the matrix
\[A=\begin{bmatrix}
\frac{2}{7} & \frac{3}{7} & \frac{6}{7} \\[6 pt]
\frac{6}{7} &\frac{2}{7} &-\frac{3}{7} \\[6pt]
-\frac{3}{7} & \frac{6}{7} & -\frac{2}{7}
\end{bmatrix}.\]
Hint.
You may use the augmented matrix […]
Inner Product, Norm, and Orthogonal Vectors
Let $\mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3$ are vectors in $\R^n$. Suppose that vectors $\mathbf{u}_1$, $\mathbf{u}_2$ are orthogonal and the norm of $\mathbf{u}_2$ is $4$ and $\mathbf{u}_2^{\trans}\mathbf{u}_3=7$. Find the value of the real number $a$ in […]
Eigenvalues of a Hermitian Matrix are Real Numbers
Show that eigenvalues of a Hermitian matrix $A$ are real numbers.
(The Ohio State University Linear Algebra Exam Problem)
We give two proofs. These two proofs are essentially the same.
The second proof is a bit simpler and concise compared to the first one.
[…]
Equivalent Conditions to be a Unitary Matrix
A complex matrix is called unitary if $\overline{A}^{\trans} A=I$.
The inner product $(\mathbf{x}, \mathbf{y})$ of complex vector $\mathbf{x}$, $\mathbf{y}$ is defined by $(\mathbf{x}, \mathbf{y}):=\overline{\mathbf{x}}^{\trans} \mathbf{y}$. The length of a complex vector […]
Normalize Lengths to Obtain an Orthonormal Basis
Let
\[
\mathbf{v}_{1}
=
\begin{bmatrix}
1 \\ 1
\end{bmatrix}
,\;
\mathbf{v}_{2}
=
\begin{bmatrix}
1 \\ -1
\end{bmatrix}
.
\]
Let $V=\Span(\mathbf{v}_{1},\mathbf{v}_{2})$. Do $\mathbf{v}_{1}$ and $\mathbf{v}_{2}$ form an orthonormal basis for $V$?
If […]
Suppose that $T: \R^2 \to \R^3$ is a linear transformation satisfying \[T\left(\, \begin{bmatrix} 1 \\ 2 \end{bmatrix}\,\right)=\begin{bmatrix} 3 \\ 4...