If the Nullity of a Linear Transformation is Zero, then Linearly Independent Vectors are Mapped to Linearly Independent Vectors

Linear Transformation problems and solutions

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 $\R^m$.

 
LoadingAdd to solve later

Proof.

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 $\R^m$.


LoadingAdd to solve later

More from my site

  • Linear Transformation to 1-Dimensional Vector Space and Its KernelLinear 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 SameDimension 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 […]
  • Range, Null Space, Rank, and Nullity of a Linear Transformation from $\R^2$ to $\R^3$Range, Null Space, Rank, and Nullity of a Linear Transformation from $\R^2$ to $\R^3$ Define the map $T:\R^2 \to \R^3$ by $T \left ( \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}\right )=\begin{bmatrix} x_1-x_2 \\ x_1+x_2 \\ x_2 \end{bmatrix}$. (a) Show that $T$ is a linear transformation. (b) Find a matrix $A$ such that […]
  • Subspace Spanned By Cosine and Sine FunctionsSubspace 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 $T (f) (x) = x f(x)$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 SubspacesA 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 SubspaceEvery 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 DiagonalizableIdempotent 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 […]

You may also like...

Leave a Reply

Your email address will not be published. Required fields are marked *

More in Linear Algebra
Inverse Matrices Problems and Solutions
Find All Values of $x$ such that the Matrix is Invertible

Given any constants $a,b,c$ where $a\neq 0$, find all values of $x$ such that the matrix $A$ is invertible if...

Close