A Linear Transformation Maps the Zero Vector to the Zero Vector

Ohio State University exam problems and solutions in mathematics

Problem 5

Let $T : \mathbb{R}^n \to \mathbb{R}^m$ be a linear transformation.
Let $\mathbf{0}_n$ and $\mathbf{0}_m$ be zero vectors of $\mathbb{R}^n$ and $\mathbb{R}^m$, respectively.
Show that $T(\mathbf{0}_n)=\mathbf{0}_m$.

[The Ohio State University exam]


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  1.  Express $\mathbf{0}_n=\mathbf{0}_n + \mathbf{0}_n$ or $\mathbf{0}_n=0\cdot \mathbf{0}_n$.
  2.  Calculate $T(\mathbf{0}_n)$ using step 1 and the definition of linear transformation.

I will give three proofs.

Proof 1.

Since  $\mathbf{0}_n=\mathbf{0}_n + \mathbf{0}_n$, we  have
\[ T( \mathbf{0}_n)=T( \mathbf{0}_n + \mathbf{0}_n)=T( \mathbf{0}_n)+T( \mathbf{0}_n),\] where the second equality follows since $T$ is a linear transformation.

Subtracting $T( \mathbf{0}_n)$ from both sides of the equality, we obtain $\mathbf{0}_m=T(\mathbf{0}_n)$.
Note that $\mathbf{0}_m=T( \mathbf{0}_n)-T( \mathbf{0}_n)$ since $T(\mathbf{0}_n)$ is a vector in $\mathbb{R}^m$.

Proof 2.

Observe that we have $0\cdot \mathbf{0}_n=\mathbf{0}_n$. (This is a scalar multiplication of the scalar $0$ and the vector $\mathbf{0}_n$.

Now we have

\[ T(\mathbf{0}_n)=T(0 \cdot \mathbf{0}_n)=0\cdot T(\mathbf{0}_n)=\mathbf{0}_m.\]

Here we used the one of the properties of the linear transformation $T$ in the second equality.

Proof 3.

Note that $\mathbf{0}_n=\mathbf{0}_n – \mathbf{0}_n$.
Thus we have
\[ T( \mathbf{0}_n)=T( \mathbf{0}_n – \mathbf{0}_n)=T( \mathbf{0}_n)-T( \mathbf{0}_n)=\mathbf{0}_m,\] where we used the linearity of $T$ in the second equality.
In the last equality, note that the vector $T(\mathbf{0}_n)$ is $m$-dimensional vector.

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2 Responses

  1. 08/15/2016

    […] that any linear transformation maps the zero vector to the zero vector. (See A linear transformation maps the zero vector to the zero vector for a proof of this […]

  2. 11/15/2016

    […] To show that $mathbf{x}, Tmathbf{x}, T^2mathbf{x}$ form a basis, it suffices to prove that these vectors are linearly independent since the dimension of the vector space $R^3$ is three and any three linearly independent vectors form a basis in such a vector space. (For a different proof, see the post A linear transformation maps the zero vector to the zero vector.) […]

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