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.

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 $\mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3$ 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 $\mathbf{u_1}=\mathbf{u_2}+a\mathbf{u}_3$.

(The Ohio State University, Linear Algebra Exam Problem)
 

Hint.

Recall the following definitions.

• The inner product (dot product) of two vectors $\mathbf{v}_1, \mathbf{v}_2$ is defined to be
  \[\mathbf{v}_1\cdot \mathbf{v}_2 :=\mathbf{v}^{\trans}_1\mathbf{v}_2.\]
• Two vectors $\mathbf{v}_1, \mathbf{v}_2$ are orthogonal if the inner product
  \[\mathbf{v}_1\cdot \mathbf{v}_2=0.\]
• The norm (length, magnitude) of a vector $\mathbf{v}$ is defined to be
  \[||\mathbf{v}||=\sqrt{\mathbf{v}\cdot \mathbf{v}}.\]

Solution.

We first express the given conditions in term of inner products (dot products).
Since $\mathbf{u}_1$ and $\mathbf{u}_2$ are orthogonal, the inner product
\[\mathbf{u}_2\cdot \mathbf{u}_1=\mathbf{u}_1 \cdot \mathbf{u}_2=0. \tag{a}\]

Also, since the norm of $\mathbf{u}_2$ is $4$, we obtain
\[\mathbf{u}_2\cdot\mathbf{u}_2=||\mathbf{u}_2||^2=16. \tag{b}\] The last condition can be written as
\[\mathbf{u}_2\cdot \mathbf{u}_3=\mathbf{u}_2^{\trans}\mathbf{u}_3=7 \tag{c}.\]

Now we compute the inner product $\mathbf{u}_2$ and $\mathbf{u}_1$.
\begin{align*}
0\stackrel{(a)}{=}&\mathbf{u}_2\cdot \mathbf{u}_1 =\mathbf{u}_2\cdot (\mathbf{u}_2+a\mathbf{u}_3)\\
&=\mathbf{u}_2\cdot \mathbf{u}_2+a\mathbf{u}_2\cdot \mathbf{u}_3\\
&\stackrel{(b), (c)}{=}16+7a.
\end{align*}
Therefore, solving this we obtain
\[a=-\frac{16}{7}.\] Click here if solved 43

The post Inner Product, Norm, and Orthogonal Vectors 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.