True or False Problems of Vector Spaces and Linear Transformations

Ohio State University exam problems and solutions in mathematics

Problem 364

These are True or False problems.
For each of the following statements, determine if it contains a wrong information or not.

  1. Let $A$ be a $5\times 3$ matrix. Then the range of $A$ is a subspace in $\R^3$.
  2. The function $f(x)=x^2+1$ is not in the vector space $C[-1,1]$ because $f(0)=1\neq 0$.
  3. Since we have $\sin(x+y)=\sin(x)+\sin(y)$, the function $\sin(x)$ is a linear transformation.
  4. The set
    \[\left\{\, \begin{bmatrix}
    1 \\
    0 \\
    0
    \end{bmatrix}, \begin{bmatrix}
    0 \\
    1 \\
    1
    \end{bmatrix} \,\right\}\] is an orthonormal set.

(Linear Algebra Exam Problem, The Ohio State University)

 
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Solution.

(1) True or False? Let $A$ be a $5\times 3$ matrix. Then the range of $A$ is a subspace in $\R^3$.

The answer is “False”. The definition of the range of the $5 \times 3$ matrix $A$ is
\[ \calR(A)=\{\mathbf{y}\in \R^5 \mid A\mathbf{x}=\mathbf{y} \text{ for some $\mathbf{x} \in \R^3$}\}.\] Note that to make sense the matrix product $A\mathbf{x}$, the size of the vector $\mathbf{x}$ must be $3$-dimensional because $A$ is $5\times 3$. Hence $\mathbf{y}=A\mathbf{x}$ is a $5$-dimensional vector, and thus the range $\calR(A)$ is a subspace of $\R^5$.
 

(2) True or False? The function $f(x)=x^2+1$ is not in the vector space $C[-1,1]$ because $f(0)=1\neq 0$.

The answer is “False”. The vector space $C[-1, 1]$ consists of all continuos functions defined on the interval $[-1, 1]$. Since $f(x)=x^2+1$ is a continuos function defined on $[-1, 1]$, it is in the vector space $C[-1, 1]$. The condition $f(0)=1\neq 0$ is irrelevant.
 

(3) True or False? Since we have $\sin(x+y)=\sin(x)+\sin(y)$, the function $\sin(x)$ is a linear transformation.

The answer is “False”. First of all $\sin(x+y)\neq \sin(x)+\sin(y)$. For example, let $x=y=\pi/2$. Then
\[\sin\left(\,\frac{\pi}{2}+\frac{\pi}{2}\,\right)=\sin\left(\,\pi \,\right)=0\] and
\[\sin\left(\, \frac{\pi}{2} \,\right)+\sin\left(\, \frac{\pi}{2} \,\right)=1+1=2.\] Hence $\sin(x)$ is not a linear transformation.
 

(4) True or False? The given set is an orthonormal set.

The answer is “False”. The dot product of these vectors is
\[\begin{bmatrix}
1 \\
0 \\
0
\end{bmatrix}\cdot \begin{bmatrix}
0 \\
1 \\
1
\end{bmatrix}=1\cdot 0+ 0\cdot 1 +0\cdot 1=0.\] Thus, the vectors are orthogonal. However the length of the second vector is
\[\sqrt{0^2+1^2+1^2}=\sqrt{2},\] hence it is not the unit vector.
So the set is orthogonal, but not orthonormal set.

Linear Algebra Midterm Exam 2 Problems and Solutions


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