# Show a linear transformation is injective using the dimension theorem.

Consider the linear transformation given by L(x)=Ax where A is the matrix $$\left(\begin{matrix}1&0&3\\2&1&2&\\1&2&-5\\2&1&1\end{matrix}\right)$$

Using the dimension theorem show that L is injective

Dimension Theorem Let V be a vector space over a ﬁeld F of ﬁnite dimension and let L : V → W be a linear transformation from V to another vector space W over F. Then:

Dim(Ker(L)) + Dim(Im(L)) = Dim(V)

I know that L is injective if and only if dim(Ker(L)) = 0; how can I show that dim(Ker(L)) = 0 using the dimension theorem?

• Prove that the $\text{Dim(Im(L))}=\text{Dim(V)}$. This imply that $\text{Dim(Ker(L))}=0$ – Carlos Jiménez May 2 '17 at 2:13

Verify how many linearly independent rows has the matrix, and substract that to the total number of rows, that's the Im(L) dimension. So, from the matrix you can see Dim(V)=3. Then , if Dim(Ker[L])=3- Dim(Im[L]) = 0 $\Rightarrow$ L is inyective
Convert the given matrix, say $A$ into row Echelon form to compute the rank of $A$ which will come out to be $3$. Now by rank-nullity theorem
$Rank(A) + nullity (A) = 3$, which implies that $nullity (A) = dim (Ker A) = 0$. Note that, since $A$ is a $4\times 3$ matrix, considering it as a linear transformation, $A$ will be a map from $\mathbb{R}^3$ to $\mathbb{R}^4$.