Linear Algebra

Prove the Cauchy-Schwarz Inequality

Problem 355

Let $\mathbf{a}, \mathbf{b}$ be vectors in $\R^n$.

Prove the Cauchy-Schwarz inequality:
\[|\mathbf{a}\cdot \mathbf{b}|\leq \|\mathbf{a}\|\,\|\mathbf{b}\|.\]

Proof 1

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.

Add to solve later

More from my site

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 Orthogonal Matrices Have Length 1. Every $3\times 3$ Orthogonal Matrix Has 1 as an Eigenvalue$ Eigenvalues of Orthogonal Matrices Have Length 1. Every $3\times 3$ Orthogonal Matrix Has 1 as an Eigenvalue (a) Let $A$ be a real orthogonal $n\times n$ matrix. Prove that the length (magnitude) of each eigenvalue of $A$ is $1$. (b) Let $A$ be a real orthogonal $3\times 3$ matrix and suppose that the determinant of $A$ is $1$. Then prove that $A$ has $1$ as an […]
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 […]