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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?

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  • $\begingroup$ How do you "know" the range is $\mathbb R^4$? Also, you can try to write $T$ in the matrix-form, and reduce it by Gauss elimination, or some similar procedures. $\endgroup$
    – awllower
    Feb 6, 2016 at 10:26
  • $\begingroup$ The codomain is $\mathbf R^4$, not the range, which is generated by the column vectors of the matrix, hence it has dimension $3$. $\endgroup$
    – Bernard
    Feb 6, 2016 at 10:42

2 Answers 2

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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

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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 $det\neq0$?

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