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Let $E$ be a vector space, and $B = \{e_1,e_2,e_3\}$ a basis of $E$. Let $f$ be an endomorphism of $E$ such that: $$\begin{align}f(e_1)&= e_1 - e_3\\ f(e_2)&= e_2 - e_3\\ f(e_1 + e_2 + e_3)&= 0\end{align}$$

Being $f$ a linear map, $f(e_1 + e_2 + e_3)= 0 = f(e_1) + f(e_2) + f(e_3)$. Then, $f(e_3)= 2e_3 - e_1 - e_2$.

Now, the matrix $A$ of $f$ with respect to $B$ (given as column vectors) is:

$A = (e_1 - e_3,\; e_2 - e_3,\; 2e_3 - e_1 - e_2)$.

They ask me to determine whether $E$ is equal to the direct sum $\text{Im}(f) + \ker(f)$.

The problem is that I have been taught to calculate $\ker(f)$ by solving the system: $AX = 0$.

How can I do that without knowing its components?

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We know that $\dim(E)=\dim(\operatorname{im}(f))+\dim(\ker(f))$ which is true for any endomorphism on a finite dimensional vectorspace.
Now you already know what $\ker(f)$ is, if you look closely, since $f(e_1)$ and $f(e_2)$ are linearly independent. An easy calculation shows that $\operatorname{im}(f)\cap\ker(f)=\{0\}$. Thus $E=\operatorname{im}(f)+\ker(f)$

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So, because of your calculation, the matrix for $f $ wrt the given basis is $\Biggr(\begin {matrix} 1&0&-1\\0&1&-1\\-1&-1&2\end {matrix}\Biggr ) $. ..

Now, $imf $ is the span of the columns, which is the span of the first two columns (the third is obviously a linear combination of the first two )... and $kerf=\{(x,x,x):x\in F \} =span\{(1,1,1)\}$...

$$(1,0,-1),(0,1,-1),(1,1,1) $ clearly span E, and are linearly independent. .. thus the answer is yes. ..

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