If the Nullity of a Linear Transformation is Zero, then Linearly Independent Vectors are Mapped to Linearly Independent Vectors
Problem 722
Let $T: \R^n \to \R^m$ be a linear transformation.
Suppose that the nullity of $T$ is zero.
If $\{\mathbf{x}_1, \mathbf{x}_2,\dots, \mathbf{x}_k\}$ is a linearly independent subset of $\R^n$, then show that $\{T(\mathbf{x}_1), T(\mathbf{x}_2), \dots, T(\mathbf{x}_k) \}$ is a linearly independent subset of $\R^m$.
Suppose that we have a linear combination
\[c_1T(\mathbf{x}_1)+c_2T(\mathbf{x}_2)+\cdots+c_k T(\mathbf{x}_k)=\mathbf{0}_m,\]
where $\mathbf{0}_m$ is the $m$ dimensional zero vector in $\R^m$.
To show that the set $\{T(\mathbf{x}_1), T(\mathbf{x}_2), \dots, T(\mathbf{x}_k) \}$ is linearly independent, we need to show that $c_1=c_2=\cdots=c_k=0$.
Using the linearity of $T$, we have
\[T(c_1\mathbf{x}_1+c_2\mathbf{x}_2+\cdots+c_k \mathbf{x}_k)=\mathbf{0}_m.\]
Then the vector $c_1\mathbf{x}_1+c_2\mathbf{x}_2+\cdots+c_k \mathbf{x}_k$ is in the nullspace $\calN(T)$ of $T$. Since the nullity, which is the dimension of the nullspace, is zero, we have $\calN(T)=\{\mathbf{0}_n\}$. This yields
\[c_1\mathbf{x}_1+c_2\mathbf{x}_2+\cdots+c_k \mathbf{x}_k=\mathbf{0}_n.\]
Since the vectors $\mathbf{x}_1, \mathbf{x}_2,\dots, \mathbf{x}_k$ are linearly independent, we must have $c_1=c_2=\dots=c_k=0$ as required.
Thus we conclude that $\{T(\mathbf{x}_1), T(\mathbf{x}_2), \dots, T(\mathbf{x}_k) \}$ is a linearly independent subset of $\R^m$.
Linear Transformation to 1-Dimensional Vector Space and Its Kernel
Let $n$ be a positive integer. Let $T:\R^n \to \R$ be a non-zero linear transformation.
Prove the followings.
(a) The nullity of $T$ is $n-1$. That is, the dimension of the nullspace of $T$ is $n-1$.
(b) Let $B=\{\mathbf{v}_1, \cdots, \mathbf{v}_{n-1}\}$ be a basis of the […]
Dimension of Null Spaces of Similar Matrices are the Same
Suppose that $n\times n$ matrices $A$ and $B$ are similar.
Then show that the nullity of $A$ is equal to the nullity of $B$.
In other words, the dimension of the null space (kernel) $\calN(A)$ of $A$ is the same as the dimension of the null space $\calN(B)$ of […]
Subspace Spanned By Cosine and Sine Functions
Let $\calF[0, 2\pi]$ be the vector space of all real valued functions defined on the interval $[0, 2\pi]$.
Define the map $f:\R^2 \to \calF[0, 2\pi]$ by
\[\left(\, f\left(\, \begin{bmatrix}
\alpha \\
\beta
\end{bmatrix} \,\right) \,\right)(x):=\alpha \cos x + \beta […]
The Range and Nullspace of the Linear Transformation $T (f) (x) = x f(x)$
For an integer $n > 0$, let $\mathrm{P}_n$ be the vector space of polynomials of degree at most $n$. The set $B = \{ 1 , x , x^2 , \cdots , x^n \}$ is a basis of $\mathrm{P}_n$, called the standard basis.
Let $T : \mathrm{P}_n \rightarrow \mathrm{P}_{n+1}$ be the map defined by, […]
A Matrix Representation of a Linear Transformation and Related Subspaces
Let $T:\R^4 \to \R^3$ be a linear transformation defined by
\[ T\left (\, \begin{bmatrix}
x_1 \\
x_2 \\
x_3 \\
x_4
\end{bmatrix} \,\right) = \begin{bmatrix}
x_1+2x_2+3x_3-x_4 \\
3x_1+5x_2+8x_3-2x_4 \\
x_1+x_2+2x_3
\end{bmatrix}.\]
(a) Find a matrix $A$ such that […]
Every Plane Through the Origin in the Three Dimensional Space is a Subspace
Prove that every plane in the $3$-dimensional space $\R^3$ that passes through the origin is a subspace of $\R^3$.
Proof.
Each plane $P$ in $\R^3$ through the origin is given by the equation
\[ax+by+cz=0\]
for some real numbers $a, b, c$.
That is, the […]
Idempotent Matrices are Diagonalizable
Let $A$ be an $n\times n$ idempotent matrix, that is, $A^2=A$. Then prove that $A$ is diagonalizable.
We give three proofs of this problem. The first one proves that $\R^n$ is a direct sum of eigenspaces of $A$, hence $A$ is diagonalizable.
The second proof proves […]