# True or False Problems of Vector Spaces and Linear Transformations ## 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) Add to solve later

## 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 Add to solve later

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