Ma · Mathematics

Do All Linear Transformations Map Zero to Zero?

Published By the 9to5Science editorial team

At a glance

  1. 1. Use additivity

    Apply the defining rule T(u + v) = T(u) + T(v) with both vectors equal to zero.

  2. 2. Compare both sides

    The equation becomes T(0) = T(0) + T(0), which forces the output to be the codomain's zero vector.

  3. 3. Check the converse

    Mapping zero to zero is necessary for linearity, but it is not enough to prove a map is linear.

Yes. Every linear transformation sends the zero vector in its domain to the zero vector in its codomain. The result follows directly from the definition of linearity and does not require the vector spaces to be finite-dimensional.

The zero vectors in the two spaces may be different objects, even though both are often written as 00. To avoid confusion, write 0V0_V for the zero vector in the domain VV and 0W0_W for the zero vector in the codomain WW. Then the statement is T(0V)=0WT(0_V)=0_W.

Why a linear map sends zero to zero

A linear transformation T:V→WT:V\to W preserves addition and scalar multiplication. In particular, it preserves addition:

T(u+v)=T(u)+T(v).T(u+v)=T(u)+T(v).

Set both uu and vv equal to the domain’s zero vector. Since 0V+0V=0V0_V+0_V=0_V,

T(0V)=T(0V+0V)=T(0V)+T(0V).\begin{aligned} T(0_V) &= T(0_V+0_V) \\ &= T(0_V)+T(0_V). \end{aligned}

Subtract T(0V)T(0_V) from both sides in the vector space WW. This leaves

T(0V)=0W.T(0_V)=0_W.

There is also a short proof using scalar multiplication. Linearity implies T(0V)=T(0⋅v)=0⋅T(v)=0WT(0_V)=T(0\cdot v)=0\cdot T(v)=0_W for any v∈Vv\in V. The additivity proof works even if you do not want to choose a particular vector vv.

A concrete matrix example

Consider the linear map T:R2→R2T:\mathbb{R}^2\to\mathbb{R}^2 defined by

T(xy)=(21−13)(xy).T\begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix}2&1\\-1&3\end{pmatrix} \begin{pmatrix}x\\y\end{pmatrix}.

At the zero vector,

T(00)=(21−13)(00)=(00).T\begin{pmatrix}0\\0\end{pmatrix} = \begin{pmatrix}2&1\\-1&3\end{pmatrix} \begin{pmatrix}0\\0\end{pmatrix} = \begin{pmatrix}0\\0\end{pmatrix}.

This is true for every matrix representing a linear map: multiplying a matrix by the zero vector gives the zero vector of the appropriate output dimension.

Does mapping zero to zero prove linearity?

No. A function can send zero to zero and still fail to be linear. For example, define f:R→Rf:\mathbb{R}\to\mathbb{R} by f(x)=x2f(x)=x^2. It satisfies f(0)=0f(0)=0, but it does not preserve addition:

f(1+1)=4,f(1)+f(1)=2.f(1+1)=4,\qquad f(1)+f(1)=2.

Since these values differ, ff is not linear. To establish linearity, you must check both required properties: preservation of addition and preservation of scalar multiplication. The zero-vector property is a useful consequence and a quick way to rule out linearity when it fails.

What happens for affine maps?

An affine map often has the form F(x)=Ax+bF(x)=A x+b, where AA is a linear map and bb is a fixed vector. At the zero vector,

F(0)=b.F(0)=b.

So an affine map need not send zero to zero. For instance, the translation F(x)=x+3F(x)=x+3 on R\mathbb{R} sends 00 to 33 and is not linear. If b=0b=0, the affine formula reduces to the linear map F(x)=AxF(x)=Ax and does send zero to zero.

Common mistakes

  • Confusing the two zero vectors. The input is 0V0_V and the output is 0W0_W. They belong to different spaces, even when both are denoted by 00.
  • Treating the property as a full test. If a function sends zero to zero, that alone does not show it is linear; x2x^2 is a counterexample.
  • Assuming every transformation is linear. Translations and other affine maps are transformations, but they are not necessarily linear.
  • Using the wrong output dimension. For a map from Rn\mathbb{R}^n to Rm\mathbb{R}^m, the result is the zero vector in Rm\mathbb{R}^m.

In practice, if a proposed linear transformation sends the domain’s zero vector to a nonzero output, you can immediately conclude that it is not linear.

References

  1. Linear Algebra Done Right, Sheldon Axler
  2. Linear Algebra, Stephen H. Friedberg, Arnold J. Insel, and Lawrence E. Spence

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