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 orthogonal to $\mathbf{v}_1$ and $\mathbf{v}_3$, and satisfying
\[ \mathbf{v}_2 \cdot \mathbf{v}_4 = -3 . \]

Calculate the following expressions:

(a) $\mathbf{v}_1 \cdot \mathbf{v}_2 $.

(b) $\mathbf{v}_3 \cdot \mathbf{v}_4$.

(c) $( 2 \mathbf{v}_1 + 3 \mathbf{v}_2 – \mathbf{v}_3 ) \cdot \mathbf{v}_4 $.

(d) $\| \mathbf{v}_1 \| , \, \| \mathbf{v}_2 \| , \, \| \mathbf{v}_3 \| $.

(e) What is the distance between $\mathbf{v}_1$ and $\mathbf{v}_2$?

Solution.

(a) $\mathbf{v}_1 \cdot \mathbf{v}_2 $.

\[ \mathbf{v}_1 \cdot \mathbf{v}_2 = \begin{bmatrix} -1 & 0 & 2 \end{bmatrix} \begin{bmatrix} 0 \\ 2 \\ -3 \end{bmatrix} = -6 . \]

(b) $\mathbf{v}_3 \cdot \mathbf{v}_4$.

We are given that $\mathbf{v}_3$ and $\mathbf{v}_4$ are orthogonal vectors, thus
\[ \mathbf{v}_3 \cdot \mathbf{v}_4 = 0 . \]

(c) $( 2 \mathbf{v}_1 + 3 \mathbf{v}_2 – \mathbf{v}_3 ) \cdot \mathbf{v}_4 $.

First, distribute the dot product over the sum:
\[ ( 2 \mathbf{v}_1 + 3 \mathbf{v}_2 – \mathbf{v}_3 ) \cdot \mathbf{v}_4 = 2 \mathbf{v}_1 \cdot \mathbf{v}_4 + 3 \mathbf{v}_2 \cdot \mathbf{v}_4 – \mathbf{v}_3 \cdot \mathbf{v}_4 . \]

Next we use the given value for $\mathbf{v}_2 \cdot \mathbf{v}_4$, along with the given facts that $\mathbf{v}_4$ is orthogonal to both $\mathbf{v}_1$ and $\mathbf{v}_3$:
\begin{align*}
&2 \mathbf{v}_1 \cdot \mathbf{v}_4 + 3 \mathbf{v}_2 \cdot \mathbf{v}_4 – \mathbf{v}_3 \cdot \mathbf{v}_4 \\
&=2\cdot 0 +3 \cdot (-3)-0 =-9.
\end{align*}

(d) $\| \mathbf{v}_1 \| , \, \| \mathbf{v}_2 \| , \, \| \mathbf{v}_3 \| $.

The length of a general vector $\mathbf{w}$ is $\|\mathbf{w}\|:=\sqrt{ \mathbf{w}^{\trans} \mathbf{w} }$. Thus,
\[ \| \mathbf{v}_1 \| \, = \, \sqrt{ (-1)^2+0^2+2^2} \, = \, \sqrt{5} , \] \[ \| \mathbf{v}_2 \| \, = \, \sqrt{ 0^2+2^2+(-3)^2} \, = \, \sqrt{13} , \] \[ \| \mathbf{v}_3 \| \, = \, \sqrt{ 2^2+2^2+3^2 } \, = \, \sqrt{17} . \]

(e) What is the distance between $\mathbf{v}_1$ and $\mathbf{v}_2$?

The distance between the two vectors is defined to be $ \| \mathbf{v}_1 – \mathbf{v}_2 \| $. First we calculate
\[ \mathbf{v}_1 – \mathbf{v}_2 \, = \, \begin{bmatrix} -1 \\ 0 \\ 2 \end{bmatrix} – \begin{bmatrix} 0 \\ 2 \\ -3 \end{bmatrix} \, = \, \begin{bmatrix} -1 \\ -2 \\ 5 \end{bmatrix} . \]

Thus,
\[ \| \mathbf{v}_1 – \mathbf{v}_2 \| = \sqrt{ 1 + 4 + 25 } = \sqrt{30} . \] Click here if solved 18

The post Inner Products, Lengths, and Distances of 3-Dimensional Real Vectors first appeared on Problems in Mathematics.

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.

Proof 1.

Let $\lambda$ be an arbitrary eigenvalue of a Hermitian matrix $A$ and let $\mathbf{x}$ be an eigenvector corresponding to the eigenvalue $\lambda$.
Then we have
\[ A\mathbf{x}= \lambda \mathbf{x}. \tag{*}\] Multiplying by $\bar{\mathbf{x}}^{\trans}$ from the left, we obtain
\begin{align*}
\bar{\mathbf{x}}^{\trans}(A\mathbf{x}) &= \bar{\mathbf{x}}^{\trans}(\lambda \mathbf{x})\\
&=\lambda \bar{\mathbf{x}}^{\trans}\mathbf{x}\\&
=\lambda ||\mathbf{x}||.
\end{align*}

We also have
\begin{align*}
\bar{\mathbf{x}}^{\trans}(A\mathbf{x})=(A\mathbf{x})^{\trans} \bar{\mathbf{x}}=\mathbf{x}^{\trans}A^{\trans}\bar{\mathbf{x}}.
\end{align*}
The first equality follows because the dot product $\mathbf{u}\cdot \mathbf{v}$ of vectors $\mathbf{u}, \mathbf{v}$ is commutative.

That is, we have
\[\mathbf{u}\cdot \mathbf{v}=\mathbf{u}^{\trans}\mathbf{v}=\mathbf{v}^{\trans}\mathbf{u}=\mathbf{v}\cdot\mathbf{u}.\] We applied this fact with $\mathbf{u}=\bar{\mathbf{x}}$ and $\mathbf{v}=A\mathbf{x}$.

Thus we obtain
\[\mathbf{x}^{\trans}A^{\trans}\bar{\mathbf{x}}=\lambda ||\mathbf{x}||.\] Taking the complex conjugate of this equality, we have
\[\bar{\mathbf{x}}^{\trans}\bar{A}^{\trans}\mathbf{x}=\bar{\lambda} ||\mathbf{x}||. \tag{**}\] (Note that $\bar{\bar{\mathbf{x}}}=\mathbf{x}$. Also $\overline{||\mathbf{x}||}=||\mathbf{x}||$ because $||\mathbf{x}||$ is a real number.)

Since the matrix $A$ is Hermitian, we have $\bar{A}^{\trans}=A$.

This yields that
\begin{align*}
\bar{\lambda} ||\mathbf{x}|| &\stackrel{(**)}{=} \bar{\mathbf{x}}^{\trans}A\mathbf{x}\\
& \stackrel{(*)}{=}\bar{\mathbf{x}}^{\trans} \lambda \mathbf{x}\\
&=\lambda ||\mathbf{x}||.
\end{align*}
Recall that $\mathbf{x}$ is an eigenvector, hence $\mathbf{x}$ is not the zero vector and the length $||\mathbf{x}||\neq 0$.

Therefore, we divide by the length $||\mathbf{x}||$ and get
\[\lambda=\bar{\lambda}.\]

It follows from this that the eigenvalue $\lambda$ is a real number.
Since $\lambda$ is an arbitrary eigenvalue of $A$, we conclude that all the eigenvalues of the Hermitian matrix $A$ are real numbers.

Proof 2.

Let $\lambda$ be an arbitrary eigenvalue of a Hermitian matrix $A$ and let $\mathbf{x}$ be an eigenvector corresponding to the eigenvalue $\lambda$.
Then we have
\[ A\mathbf{x}= \lambda \mathbf{x}. \tag{*}\]

Multiplying by $\bar{\mathbf{x}}^{\trans}$ from the left, we obtain
\begin{align*}
\bar{\mathbf{x}}^{\trans}(A\mathbf{x}) &= \bar{\mathbf{x}}^{\trans}(\lambda \mathbf{x})\\
&=\lambda \bar{\mathbf{x}}^{\trans}\mathbf{x}\\&
=\lambda ||\mathbf{x}||.
\end{align*}

Now we take the conjugate transpose of both sides and get
\[\bar{\mathbf{x}}^{\trans}\bar{A}^{\trans}\mathbf{x}=\bar{\lambda}||\mathbf{x}||. \tag{***}\] Since $A$ is a Hermitian matrix, we have $\bar{A}^{\trans}=A$.

Then the left hand side becomes
\begin{align*}
\bar{\mathbf{x}}^{\trans}\bar{A}^{\trans}\mathbf{x}&=\bar{\mathbf{x}}^{\trans}A\mathbf{x}\\
& \stackrel{(*)}{=} \bar{\mathbf{x}}^{\trans}\lambda\mathbf{x}\\
&=\lambda ||\mathbf{x}||. \tag{****}
\end{align*}

Therefore comparing (***) and (****) we obtain
\[\lambda ||\mathbf{x}||=\bar{\lambda}||\mathbf{x}||.\] Since $\mathbf{x}$ is an eigenvector, it is not the zero vector and the length $||\mathbf{x}||\neq 0$.

Dividing by the length $||\mathbf{x}||$, we obtain $\lambda=\bar{\lambda}$ and this implies that $\lambda$ is a real number.
Since $\lambda$ is an arbitrary eigenvalue of $A$, we conclude that every eigenvalue of the Hermitian matrix $A$ is a real number.

Corollary

Every real symmetric matrix is Hermitian. Thus, as a corollary of the problem we obtain the following fact:

Related Question.

Note that the inequality makes sense because eigenvalues of $A$ are real by Corollary.

For a proof of this problem, see the post “Inequality about Eigenvalue of a Real Symmetric Matrix“.

The Ohio State University Linear Algebra Exam Problems and Solutions

Check out more problems from the Ohio State University Linear Algebra Exams ↴
The Ohio State University Linear Algebra.

More Eigenvalue and Eigenvector Problems

Problems about eigenvalues and eigenvectors are collected on the page:

Eigenvectors and Eigenspaces

The post Eigenvalues of a Hermitian Matrix are Real Numbers first appeared on Problems in Mathematics.

Let $A_1, A_2, \dots, A_m$ be $n\times n$ Hermitian matrices. Show that if
\[A_1^2+A_2^2+\cdots+A_m^2=\calO,\] where $\calO$ is the $n \times n$ zero matrix, then we have $A_i=\calO$ for each $i=1,2, \dots, m$.

Hint.

Recall that a complex matrix $A$ is Hermitian if the conjugate transpose of $A$ is $A$ itself.
Namely, $A$ is Hermitian if
\[\bar{A}^{\trans}=A.\]

We also use the length of a vector in the proof below.
Let $\mathbf{v}$ be $n$-dimensional complex vector. Then the length of $\mathbf{v}$ is defined to be
\[\|\mathbf{v}\|=\sqrt{\bar{\mathbf{v}}^{\trans}\mathbf{v}}.\] The length of a complex vector $\mathbf{v}$ is a non-negative real number.

The length is also called norm or magnitude.

Proof.

Let $\mathbf{x}$ be an $n$-dimensional vector, that is, $\mathbf{x}\in \R^n$.

Then for each $i$, we have
\[\bar{\mathbf{x}}^{\trans}A_i^2\mathbf{x}=\bar{\mathbf{x}}^{\trans}\bar{A}_i^{\trans}A_i\mathbf{x}=(\overline{A_i\mathbf{x}})^{\trans}(A_i\mathbf{x})=\|A_i\mathbf{x}\|^2\geq 0.\] Here, the first equality follows from the definition of a Hermitian matrix.

Now we compute
\begin{align*}
0&=\bar{\mathbf{x}}^{\trans}\calO \mathbf{x}=\bar{\mathbf{x}}^{\trans}(A_1^2+A_2^2+\cdots+A_m^2) \mathbf{x}\\
&=\bar{\mathbf{x}}^{\trans}A_1^2\mathbf{x}+\bar{\mathbf{x}}^{\trans}A_2^2\mathbf{x}+\cdots+\bar{\mathbf{x}}^{\trans}A_m^2 \mathbf{x}\\
&=\|A_1\mathbf{x}\|^2+\|A_2\mathbf{x}\|^2+\cdots +\|A_m\mathbf{x}\|^2.
\end{align*}

Since each length $\|A_i\mathbf{x}\|$ is a non-negative real number, this implies that we have $A_i\mathbf{x}=\mathbf{0}$ for all $\mathbf{x \in \R^n}$. Hence we must have $A_i=\calO$ for each $i=1,2,\dots, m$.

The post Sum of Squares of Hermitian Matrices is Zero, then Hermitian Matrices Are All Zero first appeared on Problems in Mathematics.