For this problem, use the complex vectors
\[ \mathbf{w}_1 = \begin{bmatrix} 1 + i \\ 1 – i \\ 0 \end{bmatrix} , \, \mathbf{w}_2 = \begin{bmatrix} -i \\ 0 \\ 2 – i \end{bmatrix} , \, \mathbf{w}_3 = \begin{bmatrix} 2+i \\ 1 – 3i \\ 2i \end{bmatrix} . \]

Suppose $\mathbf{w}_4$ is another complex vector which is orthogonal to both $\mathbf{w}_2$ and $\mathbf{w}_3$, and satisfies $\mathbf{w}_1 \cdot \mathbf{w}_4 = 2i$ and $\| \mathbf{w}_4 \| = 3$.

Calculate the following expressions:

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

(b) $ \mathbf{w}_1 \cdot \mathbf{w}_3 $.

(c) $((2+i)\mathbf{w}_1 – (1+i)\mathbf{w}_2 ) \cdot \mathbf{w}_4$.

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

(e) $\| 3 \mathbf{w}_4 \|$.

(f) What is the distance between $\mathbf{w}_2$ and $\mathbf{w}_3$?

Solution.

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

\[ \mathbf{w}_1 \cdot \mathbf{w}_2 = \begin{bmatrix} 1+i & 1-i & 0 \end{bmatrix} \begin{bmatrix} -i \\ 0 \\ 2-i \end{bmatrix} = (1+i)(-i) + 0 + 0 = 1 – i . \]

(b) $ \mathbf{w}_1 \cdot \mathbf{w}_3 $.

\begin{align*} \mathbf{w}_1 \cdot \mathbf{w}_3 &= \begin{bmatrix} 1+i & 1-i & 0 \end{bmatrix} \begin{bmatrix} 2+i \\ 1-3i \\ 2i \end{bmatrix} \\ &= (1+i)(2+i) + (1-i)(1-3i) + 0 \\ &= (1 + 3i) + (-2 – 4i) \\ &= -1 – i . \end{align*}

(c) $((2+i)\mathbf{w}_1 – (1+i)\mathbf{w}_2 ) \cdot \mathbf{w}_4$.

\begin{align*} ((2+i)\mathbf{w}_1 – (1+i)\mathbf{w}_2 ) \cdot \mathbf{w}_4 &= (2+i)( \mathbf{w}_1 \cdot \mathbf{w}_4) – (1+i) ( \mathbf{w}_2 \cdot \mathbf{w}_4 ) \\
&= (2+i) ( 2i ) – (1+i)(0) \\
&= -2 + 4i \end{align*}

Note that $\mathbf{w}_2 \cdot \mathbf{w}_4=0$ because these vectors are orthogonal.

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

For an arbitrary complex vector $\mathbf{v}$, its length is defined to be
\[ \| \mathbf{v} \| = \sqrt{ \overline{\mathbf{v}}^\trans \mathbf{v} } . \]

Thus,
\[ \| \mathbf{w}_1 \| \, = \, \sqrt{ (1-i)(1+i) + (1+i)(1-i) + 0 } = \sqrt{ 2 + 2} = \sqrt{4} , \] \[ \| \mathbf{w}_2 \| \, = \, \sqrt{ (i)(-i) + 0 + (2+i)(2-i) } = \sqrt{1 + 5} = \sqrt{6} , \] \[ \| \mathbf{w}_3 \| \, = \, \sqrt{ (2-i)(2+i) + (1+3i)(1-3i) + (-2i)(2i) } = \sqrt{ 5 + 10 + 4} = \sqrt{19} . \]

(e) $\| 3 \mathbf{w}_4 \|$.

$ \| 3 \mathbf{w}_4 \| = 3 \| \mathbf{w}_4 \| = 3\cdot 3=9 $ .

(f) What is the distance between $\mathbf{w}_2$ and $\mathbf{w}_3$?

The distance between these vectors is given by $\| \mathbf{w}_2 – \mathbf{w}_3 \|$. First we calculate this difference:
\[ \mathbf{w}_2 – \mathbf{w}_3 \, = \, \begin{bmatrix} -i \\ 0 \\ 2 – i \end{bmatrix} – \begin{bmatrix} 2+i \\ 1 – 3i \\ 2i \end{bmatrix} \, = \, \begin{bmatrix} -2 – 2i \\ -1 + 3i \\ 2 – 3i \end{bmatrix} . \]

Now the length of the complex vector is defined to be
\begin{align*}
\| \mathbf{w}_2 – \mathbf{w}_3 \| &= \sqrt{ \left( \overline{ \mathbf{w}_2 – \mathbf{w}_3 } \right)^{\trans} \left( \mathbf{w}_2 – \mathbf{w}_3 \right) } \\[6pt] &= \sqrt{ \begin{bmatrix} -2 + 2i & -1 – 3i & 2 + 3i \end{bmatrix} \begin{bmatrix} -2 – 2i \\ -1 + 3i \\ 2 – 3i \end{bmatrix} } \\[6pt] &= \sqrt{ (-2+2i)(-2-2i) + (-1-3i)(-1+3i) + (2+3i)(2-3i) } \\[6pt] &= \sqrt{ 8 + 10 + 13 } \\[6pt] &= \sqrt{ 31} \end{align*}

The post Dot Product, Lengths, and Distances of Complex Vectors first appeared on Problems in Mathematics.

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.

Let $\mathbf{v}$ be an $n \times 1$ column vector.

Prove that $\mathbf{v}^\trans \mathbf{v} = 0$ if and only if $\mathbf{v}$ is the zero vector $\mathbf{0}$.

Proof.

Let $\mathbf{v} = \begin{bmatrix} v_1 \\ v_2 \\ \vdots \\ v_n \end{bmatrix} $.

Then we have
\[\mathbf{v}^\trans \mathbf{v} = \begin{bmatrix} v_1 & v_2 & \cdots & v_n \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \\ \vdots \\ v_n \end{bmatrix} = \sum_{i=1}^n \mathbf{v}_i^2 . \] Because each $v_i^2$ is non-negative, this sum is $0$ if and only if $v_i = 0$ for each $i$. In this case, $\mathbf{v}$ is the zero vector.

Comment.

Recall that the the length of the vector $\mathbf{v}\in \R^n$ is defined to be
\[\|\mathbf{v}\| :=\sqrt{\mathbf{v}^{\trans} \mathbf{v}}.\]

The problem implies that the length of a vector is $0$ if and only if the vector is the zero vector.

The post The Length of a Vector is Zero if and only if the Vector is the Zero Vector first appeared on Problems in Mathematics.

Let $A$ and $B$ be $n\times n$ skew-symmetric matrices. Namely $A^{\trans}=-A$ and $B^{\trans}=-B$.

(a) Prove that $A+B$ is skew-symmetric.

(b) Prove that $cA$ is skew-symmetric for any scalar $c$.

(c) Let $P$ be an $m\times n$ matrix. Prove that $P^{\trans}AP$ is skew-symmetric.

(d) Suppose that $A$ is real skew-symmetric. Prove that $iA$ is an Hermitian matrix.

(e) Prove that if $AB=-BA$, then $AB$ is a skew-symmetric matrix.

(f) Let $\mathbf{v}$ be an $n$-dimensional column vecotor. Prove that $\mathbf{v}^{\trans}A\mathbf{v}=0$.

(g) Suppose that $A$ is a real skew-symmetric matrix and $A^2\mathbf{v}=\mathbf{0}$ for some vector $\mathbf{v}\in \R^n$. Then prove that $A\mathbf{v}=\mathbf{0}$.

Proof.

(a) Prove that $A+B$ is skew-symmetric.

We have
\begin{align*}
(A+B)^{\trans}=A^{\trans}+B^{\trans}=(-A)+(-B)=-(A+B).
\end{align*}
Hence $A+B$ is skew-symmetric.

(b) Prove that $cA$ is skew-symmetric for any scalar $c$.

We compute
\begin{align*}
(cA)^{\trans}=cA^{\trans}=c(-A)=-cA.
\end{align*}
Thus, $cA$ is skew-symmetric.

(c) Let $P$ be an $m\times n$ matrix. Prove that $P^{\trans}AP$ is skew-symmetric.

Using the properties of transpose, we have
\begin{align*}
(P^{\trans}AP)^{\trans}&=P^{\trans}A^{\trans}(P^{\trans})^{\trans}=P^{\trans}A^{\trans}P\\
&=P^{\trans}(-A)P=-(P^{\trans}AP).
\end{align*}
This implies that $P^{\trans}AP$ is skew-symmetric.

(d) Suppose that $A$ is real skew-symmetric. Prove that $iA$ is an Hermitian matrix.

Note that since $A$ is real, we have $\bar{A}=A$.
Then we have
\begin{align*}
(\overline{iA})^{\trans}=(\bar{i}\bar{A})^{\trans}=(-iA)^{\trans}=(-i)A^{\trans}=(-i)(-A)=iA.
\end{align*}
It follows that $iA$ is Hermitian.

(e) Prove that if $AB=-BA$, then $AB$ is a skew-symmetric matrix.

We calculate
\begin{align*}
(AB)^{\trans}&=B^{\trans}A^{\trans}=(-B)(-A)\\
&=BA=-AB,
\end{align*}
where the last step follows from the assumption $AB=-BA$.
This proves that $AB$ is skew-symmetric.

(f) Let $\mathbf{v}$ be an $n$-dimensional column vecotor. Prove that $\mathbf{v}^{\trans}A\mathbf{v}=0$.

Observe that $\mathbf{v}^{\trans}A\mathbf{v}$ is a $1\times 1$ matrix, or just a number.
So we have
\begin{align*}
\mathbf{v}^{\trans}A\mathbf{v}&=(\mathbf{v}^{\trans}A\mathbf{v})^{\trans}=\mathbf{v}^{\trans}A^{\trans}(\mathbf{v}^{\trans})^{\trans}\\
&=\mathbf{v}^{\trans}A^{\trans}\mathbf{v}=\mathbf{v}^{\trans}(-A)\mathbf{v}=-(\mathbf{v}^{\trans}A\mathbf{v}).
\end{align*}
This yields that $2\mathbf{v}^{\trans}A\mathbf{v}=0$, and hence $\mathbf{v}^{\trans}A\mathbf{v}=0$.

(g) Suppose that $A$ is a real skew-symmetric matrix and $A^2\mathbf{v}=\mathbf{0}$ for some vector $\mathbf{v}\in \R^n$. Then prove that $A\mathbf{v}=\mathbf{0}$.

Let us compute the length of the vector $A\mathbf{v}$.
We have
\begin{align*}
\|A\mathbf{v}\|&=(A\mathbf{v})^{\trans}(A\mathbf{v})=\mathbf{v}^{\trans}A^{\trans}A\mathbf{v}\\
&=\mathbf{v}^{\trans}(-A)A\mathbf{v}=-\mathbf{v}^{\trans}A^2\mathbf{v}\\
&=-\mathbf{v}\mathbf{0} &&\text{by assumption}\\
&=0.
\end{align*}
Since the length $\|A\mathbf{v}\|=0$, we conclude that $A\mathbf{v}=\mathbf{0}$.

The post 7 Problems on Skew-Symmetric Matrices first appeared on Problems in Mathematics.

Let $\mathbf{v}$ be a vector in an inner product space $V$ over $\R$.
Suppose that $\{\mathbf{u}_1, \dots, \mathbf{u}_n\}$ is an orthonormal basis of $V$.
Let $\theta_i$ be the angle between $\mathbf{v}$ and $\mathbf{u}_i$ for $i=1,\dots, n$.

Prove that
\[\cos ^2\theta_1+\cdots+\cos^2 \theta_n=1.\]

Definition (Angle between Vectors).

Let $\langle\mathbf{a}, \mathbf{b}\rangle$ denote the inner product of vectors $\mathbf{a}$ and $\mathbf{b}$ in $V$.

Recall that the angle $\theta$ between $\mathbf{a}$ and $\mathbf{b}$ is defined as the unique number $\theta$ between $0$ and $\pi$ satisfying
\[\cos \theta=\frac{\langle\mathbf{a}, \mathbf{b}\rangle}{\|\mathbf{a}\| \|\mathbf{b}\|}.\]

Proof.

Express the vector $\mathbf{v}$ as a linear combination of the basis vectors as
\[\mathbf{v}=a_1\mathbf{u}_1+\dots+a_n\mathbf{u}_n\] for some real numbers $a_1, \dots, a_n$.

The length of the vector $\mathbf{v}$ is given by
\[\|\mathbf{v}\|=\sqrt{a_1^2+\cdots+a_n^2}. \tag{*}\]

For each $i$, we have using the properties of the inner product
\begin{align*}
\langle \mathbf{v}, \mathbf{u}_i\rangle&=\langle a_1\mathbf{u}_1+\dots+a_n\mathbf{u}_n, \mathbf{u}_i\rangle\\
&=a_1\langle\mathbf{u}_1, \mathbf{u}_i\rangle+\cdots +a_n \langle\mathbf{u}_n, \mathbf{u}_i \rangle\\
&=a_i \tag{**}
\end{align*}
since $\langle\mathbf{u}_i, \mathbf{u}_i\rangle=1$ and $\langle\mathbf{u}_j, \mathbf{u}_i\rangle=0$ if $j\neq i$ as $\{\mathbf{u}_1, \dots, \mathbf{u}_n\}$ is orthonormal.

By definition of the angle, we have
\begin{align*}
\cos \theta_i&=\frac{\langle\mathbf{v}, \mathbf{u}_i\rangle}{\|\mathbf{v}\| \|\mathbf{u}_i\|}=\frac{\langle\mathbf{v}, \mathbf{u}_i\rangle}{\|\mathbf{v}\| } && \text{since $\|\mathbf{u}_i\|=1$.}
\end{align*}
It follows that
\begin{align*}
\cos ^2\theta_1+\cdots+\cos^2 \theta_n &=\frac{\langle\mathbf{v}, \mathbf{u}_1\rangle^2}{\|\mathbf{v}\|^2 }+\cdots+\frac{\langle\mathbf{v}, \mathbf{u}_n\rangle^2}{\|\mathbf{v}\|^2 }\\[6pt] &=\frac{1}{\|\mathbf{v}\|^2}(a_1^2+\cdots a_n^2) &&\text{by (**)}\\[6pt] &=\frac{1}{\|\mathbf{v}\|^2}\cdot \|\mathbf{v}\|^2 &&\text{by (*)}\\[6pt] &=1.
\end{align*}

Thus we obtain
\[\cos ^2\theta_1+\cdots+\cos^2 \theta_n=1\] as required.

The post The Sum of Cosine Squared in an Inner Product Space first appeared on Problems in Mathematics.

A square matrix $A$ is called idempotent if $A^2=A$.

(a) Let $\mathbf{u}$ be a vector in $\R^n$ with length $1$.
Define the matrix $P$ to be $P=\mathbf{u}\mathbf{u}^{\trans}$.

Prove that $P$ is an idempotent matrix.

(b) Suppose that $\mathbf{u}$ and $\mathbf{v}$ be unit vectors in $\R^n$ such that $\mathbf{u}$ and $\mathbf{v}$ are orthogonal.
Let $Q=\mathbf{u}\mathbf{u}^{\trans}+\mathbf{v}\mathbf{v}^{\trans}$.

Prove that $Q$ is an idempotent matrix.

(c) Prove that each nonzero vector of the form $a\mathbf{u}+b\mathbf{v}$ for some $a, b\in \R$ is an eigenvector corresponding to the eigenvalue $1$ for the matrix $Q$ in part (b).

Proof.

(a) Prove that $P=\mathbf{u}\mathbf{u}^{\trans}$ is an idempotent matrix.

The length of the vector $\mathbf{u}$ is given by
\[\|\mathbf{u}\|=\sqrt{\mathbf{u}^{\trans}\mathbf{u}}.\] Since $\|\mathbf{u}\|=1$ by assumption, it yields that
\[\mathbf{u}^{\trans}\mathbf{u}=1.\]

Let us compute $P^2$ using the associative properties of matrix multiplication.
We have
\begin{align*}
P^2&=(\mathbf{u}\mathbf{u}^{\trans})(\mathbf{u}\mathbf{u}^{\trans})\\
&=\mathbf{u}(\mathbf{u}^{\trans}\mathbf{u})\mathbf{u}^{\trans}\\
&=\mathbf{u}( 1 ) \mathbf{u}^{\trans}=\mathbf{u}\mathbf{u}^{\trans}=P.
\end{align*}

Thus, we have obtained $P^2=P$, and hence $P$ is an idempotent matrix.

(b) Prove that $Q=\mathbf{u}\mathbf{u}^{\trans}+\mathbf{v}\mathbf{v}^{\trans}$ is an idempotent matrix

Since $\mathbf{u}$ and $\mathbf{v}$ are unit vectors, we have as in part (a)
\[\mathbf{u}^{\trans}\mathbf{u}=1 \text{ and } \mathbf{v}^{\trans}\mathbf{v}=1.\] Since $\mathbf{u}$ and $\mathbf{v}$ are orthogonal, their dot (inner) product is $0$.
Thus we have
\[\mathbf{u}^{\trans}\mathbf{v}=\mathbf{v}^{\trans}\mathbf{u}=0.\]

Using these identities, we compute $Q^2$ as follows.
We have
\begin{align*}
Q^2&=(\mathbf{u}\mathbf{u}^{\trans}+\mathbf{v}\mathbf{v}^{\trans})(\mathbf{u}\mathbf{u}^{\trans}+\mathbf{v}\mathbf{v}^{\trans})\\
&=\mathbf{u}\mathbf{u}^{\trans}(\mathbf{u}\mathbf{u}^{\trans}+\mathbf{v}\mathbf{v}^{\trans})
+\mathbf{v}\mathbf{v}^{\trans}(\mathbf{u}\mathbf{u}^{\trans}+\mathbf{v}\mathbf{v}^{\trans})\\
&=\mathbf{u}\underbrace{\mathbf{u}^{\trans}\mathbf{u}}_{1}\mathbf{u}^{\trans}+\mathbf{u}\underbrace{\mathbf{u}^{\trans}\mathbf{v}}_{0}\mathbf{v}^{\trans}
+\mathbf{v}\underbrace{\mathbf{v}^{\trans}\mathbf{u}}_{0}\mathbf{u}^{\trans}+\mathbf{v}\underbrace{\mathbf{v}^{\trans}\mathbf{v}}_{1}\mathbf{v}^{\trans}\\[6pt] &=\mathbf{u}\mathbf{u}^{\trans}+\mathbf{v}\mathbf{v}^{\trans}=Q.
\end{align*}
It follows that $Q$ is an idempotent matrix.

(c) Prove that $a\mathbf{u}+b\mathbf{v}$ is an eigenvector

Let us first compute $Q\mathbf{u}$. We have
\begin{align*}
Q\mathbf{u}&=(\mathbf{u}\mathbf{u}^{\trans}+\mathbf{v}\mathbf{v}^{\trans})\mathbf{u}\\
&=\mathbf{u}\underbrace{\mathbf{u}^{\trans}\mathbf{u}}_{1}+\mathbf{v}\underbrace{\mathbf{v}^{\trans}\mathbf{u}}_{0}=\mathbf{u}.
\end{align*}

Note that $\mathbf{u}$ is a nonzero vector because it is a unit vector.
Thus, the equality $Q\mathbf{u}=\mathbf{u}$ implies that $1$ is an eigenvalue of $Q$ and $\mathbf{v}$ is a corresponding eigenvector.

Similarly, we can check that $\mathbf{v}$ is an eigenvector corresponding to the eigenvalue $1$.

Now let $a\mathbf{u}+b\mathbf{v}$ be a nonzero vector.
Then we have
\begin{align*}
Q(a\mathbf{u}+b\mathbf{v})&=aQ\mathbf{u}+bQ\mathbf{v}=a\mathbf{u}+b\mathbf{v}.
\end{align*}
It follows that $a\mathbf{u}+b\mathbf{v}$ is an eigenvector corresponding to the eigenvalue $1$.

Another way to prove (c)

Another way to see this is as follows.
As we saw above, the vectors $\mathbf{u}$ and $\mathbf{v}$ are eigenvectors corresponding to the eigenvalue $1$.
Hence $\mathbf{u}, \mathbf{v} \in E_{1}$, where $E_1$ is an eigenspace of the eigenvalue $1$

Note that $E_1$ is a vector space, hence $a\mathbf{u}+b\mathbf{v}$ is a nonzero vector in $E_{1}$.
Thus, $a\mathbf{u}+b\mathbf{v}$ is an eigenvector corresponding to the eigenvalue $1$ as well.

Related Question.

See the post ↴
Idempotent Matrix and its Eigenvalues
for solutions of this problem.

The post Unit Vectors and Idempotent Matrices first appeared on Problems in Mathematics.

Consider the matrix
\[A=\begin{bmatrix}
3/2 & 2\\
-1& -3/2
\end{bmatrix} \in M_{2\times 2}(\R).\]

(a) Find the eigenvalues and corresponding eigenvectors of $A$.

(b) Show that for $\mathbf{v}=\begin{bmatrix}
1 \\
0
\end{bmatrix}\in \R^2$, we can choose $n$ large enough so that the length $\|A^n\mathbf{v}\|$ is as small as we like.

(University of California, Berkeley, Linear Algebra Final Exam Problem)
 

Proof.

(a) Find the eigenvalues and corresponding eigenvectors of $A$.

To find the eigenvalues of $A$, we compute the characteristic polynomial $p(t)$ of $A$.
We have
\begin{align*}
p(t)&=\det(A-tI)\\
&=\begin{vmatrix}
3/2-t & 2\\
-1& -3/2-t
\end{vmatrix}\\
&=t^2-1/4.
\end{align*}
Since the eigenvalues are roots of the characteristic polynomials, the eigenvalues of $A$ are $\pm 1/2$.

Next we find the eigenvectors corresponding to eigenvalue $1/2$.
These are the solutions of $(A-\frac{1}{2}I)\mathbf{x}=\mathbf{0}$.
We have
\begin{align*}
A-\frac{1}{2}I=\begin{bmatrix}
1 & 2\\
-1& -2
\end{bmatrix}
\xrightarrow{R_2+R_1}
\begin{bmatrix}
1 & 2\\
0& 0
\end{bmatrix}.
\end{align*}
Thus, the solution $\mathbf{x}$ satisfies $x_1=-2x_2$, and the eigenvectors are
\[\mathbf{x}=x_2\begin{bmatrix}
-2 \\
1
\end{bmatrix},\] where $x_2$ is a nonzero scalar.

Similarly, we find the eigenvectors corresponding to the eigenvalue $-1/2$.
We solve $(A+\frac{1}{2}I)\mathbf{x}=\mathbf{0}$.
We have
\begin{align*}
A+\frac{1}{2}I=\begin{bmatrix}
2 & 2\\
-1& -1
\end{bmatrix}
\xrightarrow[\text{then } R_2+R_1]{\frac{1}{2}R_1}
\begin{bmatrix}
1 & 1\\
0& 0
\end{bmatrix}.
\end{align*}
Thus, we have $x_1=-x_2$, and the eigenvectors are
\[\mathbf{x}=x_2\begin{bmatrix}
-1 \\
1
\end{bmatrix},\] where $x_2$ is a nonzero scalar.

(b) We can choose $n$ large enough so that the length $\|A^n\mathbf{v}\|$ is as small as we like.

We express the vector $\mathbf{v}=\begin{bmatrix}
1 \\
0
\end{bmatrix}$ as a linear combination of eigenvectors $\begin{bmatrix}
-2 \\
1
\end{bmatrix}$ and $\begin{bmatrix}
-1 \\
1
\end{bmatrix}$ corresponding to eigenvalues $1/2$ and $-1/2$, respectively.
Let
\[\begin{bmatrix}
1 \\
0
\end{bmatrix}=c_1\begin{bmatrix}
-2 \\
1
\end{bmatrix}+c_2\begin{bmatrix}
-1 \\
1
\end{bmatrix}\] for some scalars $c_1, c_2$.
Solving this for $c_1, c_2$, we find that $c_1=-1$ and $c_2=1$, and thus we have
\[\begin{bmatrix}
1 \\
0
\end{bmatrix}=-\begin{bmatrix}
-2 \\
1
\end{bmatrix}+\begin{bmatrix}
-1 \\
1
\end{bmatrix}.\] Then for any positive integer $n$, we have
\begin{align*}
A^n\begin{bmatrix}
1 \\
0
\end{bmatrix}&=-A^n\begin{bmatrix}
-2 \\
1
\end{bmatrix}+A^n\begin{bmatrix}
-1 \\
1
\end{bmatrix}\\
&=-\left(\, \frac{1}{2} \,\right)^n\begin{bmatrix}
-2 \\
1
\end{bmatrix}+\left(\, -\frac{1}{2} \,\right)^n\begin{bmatrix}
-1 \\
1
\end{bmatrix}\\
&=\left(\, \frac{1}{2} \,\right)^n\begin{bmatrix}
2-(-1)^n \\
-1+(-1)^n
\end{bmatrix}
\end{align*}
Note that in the second equality we used the following fact: If $A\mathbf{x}=\lambda \mathbf{x}$, then $A^n\mathbf{x}=\lambda^n \mathbf{x}$.

Then the length is
\begin{align*}
\left \| A^n\begin{bmatrix}
1 \\
0
\end{bmatrix}\right \|&=\left(\, \frac{1}{2} \,\right)^n \sqrt{\left(\, 2-(-1)^n \,\right)^2+\left(\, -1+(-1)^n \,\right)^2}\\
& \leq \left(\, \frac{1}{2} \,\right)^n \sqrt{3^2+2^2}\\
&= \sqrt{13}\left(\, \frac{1}{2} \,\right)^n \to 0 \text{ as $n$ tends to infinity}.
\end{align*}
Therefore, we can choose $n$ large enough so that the length $\|A^n\mathbf{v}\|$ is as small as we wish.

The post Prove that the Length $\|A^n\mathbf{v}\|$ is As Small As We Like. first appeared on Problems in Mathematics.

Let $A=\begin{bmatrix}
1 & 0 & 1 \\
0 &1 &0
\end{bmatrix}$.

(a) Find an orthonormal basis of the null space of $A$.

(b) Find the rank of $A$.

(c) Find an orthonormal basis of the row space of $A$.

(The Ohio State University, Linear Algebra Exam Problem)
 

Solution.

First of all, note that $A$ is already in reduced row echelon form.

(a) Find an orthonormal basis of the null space of $A$.

Let us find a basis of null space of $A$.
The null space consists of the solutions of $A\mathbf{x}=0$.
Since $A$ is in reduced row echelon form, the solutions $\mathbf{x}=\begin{bmatrix}
x_1 \\
x_2 \\
x_3
\end{bmatrix}$ satisfy
\[x_1=-x_3 \text{ and } x_2=0,\] hence the general solution is
\[\mathbf{x}=\begin{bmatrix}
-x_3 \\
0 \\
x_3
\end{bmatrix}=x_3\begin{bmatrix}
-1 \\
0 \\
1
\end{bmatrix}.\] Therefore, the set
\[\left\{\, \begin{bmatrix}
-1 \\
0 \\
1
\end{bmatrix} \,\right\}\] is a basis of the null space of $A$.
Since the length of the basis vector is $\sqrt{(-1)^2+0^2+1^2}=\sqrt{2}$, it is not orthonormal basis.
Thus, we divide the vector by its length and obtain an orthonormal basis
\[\left\{\, \frac{1}{\sqrt{2}}\begin{bmatrix}
-1 \\
0 \\
1
\end{bmatrix} \,\right\}.\]  

(b) Find the rank of $A$.

From part (a), we see that the nullity of $A$ is $1$. The rank-nullity theorem says that
\[\text{rank of $A$} + \text{ nullity of $A$}=3.\] Hence the rank of $A$ is $2$.

The second way to find the rank of $A$ is to use the definition of the rank: The rank of a matrix $B$ is the number of nonzero rows in a reduced row echelon matrix that is row equivalent to $B$.
Since $A$ is in echelon form and it has two nonzero rows, the rank is $2$.

The third way to find the rank is to use the leading 1 method. By the leading 1 method, we see that the first two columns form a basis of the range, hence the rank of $A$ is $2$.
 

(c) Find an orthonormal basis of the row space of $A$.

By the row space method, the nonzero rows in reduced row echelon form a basis of the row space of $A$. Thus
\[ \left\{\, \begin{bmatrix}
1 \\
0 \\
1
\end{bmatrix}, \begin{bmatrix}
0 \\
1 \\
0
\end{bmatrix} \,\right\}\] is a basis of the row space of $A$.
Since the dot (inner) product of these two vectors is $0$, they are orthogonal.
The length of the vectors is $\sqrt{2}$ and $1$, respectively.
Hence an orthonormal basis of the row space of $A$ is
\[ \left\{\, \frac{1}{\sqrt{2}} \begin{bmatrix}
1 \\
0 \\
1
\end{bmatrix}, \begin{bmatrix}
0 \\
1 \\
0
\end{bmatrix} \,\right\}\]  

Linear Algebra Midterm Exam 2 Problems and Solutions

• True of False Problems and Solutions: True or False problems of vector spaces and linear transformations
• Problem 1 and its solution: See (7) in the post “10 examples of subsets that are not subspaces of vector spaces”
• Problem 2 and its solution: Determine whether trigonometry functions $\sin^2(x), \cos^2(x), 1$ are linearly independent or dependent
• Problem 3 and its solution (current problem): Orthonormal basis of null space and row space
• Problem 4 and its solution: Basis of span in vector space of polynomials of degree 2 or less
• Problem 5 and its solution: Determine value of linear transformation from $R^3$ to $R^2$
• Problem 6 and its solution: Rank and nullity of linear transformation from $R^3$ to $R^2$
• Problem 7 and its solution: Find matrix representation of linear transformation from $R^2$ to $R^2$
• Problem 8 and its solution: Hyperplane through origin is subspace of 4-dimensional vector space

The post Orthonormal Basis of Null Space and Row Space first appeared on Problems in Mathematics.

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 that this length is the distance between $\mathbf{a}$ and $\mathbf{b}$.)

Solution.

Recall that the length of a vector $\mathbf{x}$ is defined to be
\[\|\mathbf{x}\|=\sqrt{\mathbf{x}^{\trans}\mathbf{x}},\] where $\mathbf{x}^{\trans}$ is the transpose of $\mathbf{x}$.

Also, recall that the inner product of two vectors $\mathbf{x}, \mathbf{y}$ are commutative.
Namely we have
\[\mathbf{x}\cdot \mathbf{y}=\mathbf{x}^{\trans}\mathbf{y}=\mathbf{y}^{\trans}\mathbf{x}=\mathbf{y} \cdot \mathbf{x}.\]

Applying the second fact with given vectors $\mathbf{a}, \mathbf{b}$, we obtain
\[\mathbf{a}^{\trans}\mathbf{b}=\mathbf{b}^{\trans}\mathbf{a}= -\frac{1}{2}.\]

Now we compute $\|\mathbf{a}-\mathbf{b}\|^2$ as follows.
We have
\begin{align*}
\|\mathbf{a}-\mathbf{b}\|^2&=(\mathbf{a}-\mathbf{b})^{\trans}(\mathbf{a}-\mathbf{b}) \qquad \text{ (by definition of the length)}\\
&=(\mathbf{a}^{\trans}-\mathbf{b}^{\trans})(\mathbf{a}-\mathbf{b})\\
&=\mathbf{a}^{\trans}\mathbf{a}-\mathbf{a}^{\trans}\mathbf{b}-\mathbf{b}^{\trans}\mathbf{a}+\mathbf{b}^{\trans}\mathbf{b}\\
&=\|\mathbf{a}\|^2-\mathbf{a}^{\trans}\mathbf{b}-\mathbf{b}^{\trans}\mathbf{a}+\|\mathbf{b}\|^2\\
&=1-\left(-\frac{1}{2} \right)-\left(-\frac{1}{2} \right)+1\\
&=3.
\end{align*}

Since the length is nonnegative, we take the square root of the above equality and obtain
\[\|\mathbf{a}-\mathbf{b}\|=\sqrt{3}.\] Click here if solved 17

The post Find the Distance Between Two Vectors if the Lengths and the Dot Product are Given first appeared on Problems in Mathematics.

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 method to find the inverse matrix.
Here we give an alternative way to find the inverse matrix by noting that $A$ is an orthogonal matrix.

Recall that a matrix $B$ is orthogonal if $B^{\trans}B=B^{\trans}B=I$.
Thus, once we know $B$ is an orthogonal matrix, then the inverse matrix $B^{-1}$ is just the transpose matrix $B^{\trans}$.

Also, recall that a matrix $B$ is orthogonal if and only if the column vectors of $B$ form an orthonormal set.

Solution.

We first show that $A$ is an orthogonal matrix.
To do this, it suffices to that the column vectors form an orthonormal set.

Let
\[ \mathbf{v}_1= \begin{bmatrix}
\frac{2}{7} \\[6 pt] \frac{6}{7} \\[6pt] -\frac{3}{7}
\end{bmatrix},
\mathbf{v}_2=\begin{bmatrix}
\frac{3}{7} \\[6 pt] \frac{2}{7} \\[6pt] \frac{6}{7} \end{bmatrix},
\mathbf{v}_3=\begin{bmatrix}
\frac{6}{7} \\[6 pt] -\frac{3}{7} \\[6pt] -\frac{2}{7}
\end{bmatrix}\] be the column vectors of $A$.

Then the length of the vector $\mathbf{v}_1$ is
\[||\mathbf{v}_1||=\sqrt{(2/7)^2+(6/7)^2+(-3/7)^2}=1.\] Similarly, we have $||\mathbf{v}_2||=||\mathbf{v}_3||=1$.
Thus, column vectors are unit vectors.

The dot (inner) product of the vectors $\mathbf{v}_1$ and $\mathbf{v}_2$ is
\[\mathbf{v}_1\cdot \mathbf{v}_2=\frac{2}{7}\cdot \frac{3}{7}+\frac{6}{7}\cdot \frac{2}{7}+\left( -\frac{3}{7}\right) \cdot \frac{6}{7}=0.\] Similarly, we have
\[\mathbf{v}_1\cdot \mathbf{v}_3=0, \quad \mathbf{v}_2\cdot \mathbf{v}_3=0.\]

Therefore, the column vectors are orthogonal.
Hence the column vectors of $A$ are orthonormal, and this implies that $A$ is an orthogonal matrix. Namely, $A^{\trans}=A^{-1}$.
Thus the inverse matrix of $A$ is
\[A^{-1}=\begin{bmatrix}
\frac{2}{7} & \frac{6}{7} & -\frac{3}{7} \\[6 pt] \frac{3}{7} &\frac{2}{7} &\frac{6}{7} \\[6pt] \frac{6}{7} & -\frac{3}{7} & -\frac{2}{7}
\end{bmatrix}.\] Click here if solved 48

The post Find the Inverse Matrix of a Matrix With Fractions first appeared on Problems in Mathematics.