Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.
Sign up to join this community
Find the rank and the nullity of the following linear map $T : U \to V$ , and find bases of the kernel and image of $T$.
$U = \Bbb R^4 , V = \Bbb R^4$, $$T(α, β, γ, δ) = (α − γ, γ − δ, α − β, β − δ)$$
Attempt of the question: I have tried to find a basis of kernel, it is $(1,1,1,1)$. so $\text{nullity}(T)=1$
But then the image is $\Bbb R^4$ which means $\text{rank}(T)=4$.
$\dim(U)=4$ and $\text{rank}+\text{nullity}=5$
Where did I make the mistake?
The image is a subspace of $\;\Bbb R^4\;$ , not the whole space. Take for example the matrix of the transformation using the canonical basis and reduce it by columns:
$$\begin{pmatrix}1&0&-1&0\\0&1&0&-1\\1&-1&0&0\\0&1&-1&0\end{pmatrix}\xrightarrow{C_3+C_1}\begin{pmatrix}1&0&0&0\\0&1&0&-1\\0&-1&1&0\\0&1&-1&0\end{pmatrix}\xrightarrow{C_4+C_2}\begin{pmatrix}1&0&0&0\\0&1&0&0\\1&-1&1&-1\\0&1&-1&1\end{pmatrix}\xrightarrow{}$$
$$\xrightarrow{C_4+ C_3}\begin{pmatrix}1&0&0&0\\0&1&0&0\\1&-1&1&0\\0&1&-1&0\end{pmatrix}\xrightarrow{}$$
and you can see the columns space = image of the transformation is generated by the first three columnvectors
Consider the matrix of the transformation taking canonical basis:$$A=\begin{pmatrix}1&0&-1&0\\0&0&0&-1\\-1&1&0&0\\0&-1&0&-1\end{pmatrix}$$ and calculate the determinant of this matrix. If $\det=0$ or $det\neq0$?
