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Problem: For the matrix $A\in M_{3\times3}$, the eigenvalues are known: $Sp(A)=\{0,1,2\}$, and the corresponding eigenvectors are $u, v, w$, respectively. Determine a basis for the null space $N(A)$.

The solution just says that $N(A)=N(A-0*I)= L(u)$, but I don't understand why that is true. I understand that $u$ is an element of $N(A)$, but how do we know that it is enough to be a basis? How do we know that all elements of $N(A)$ can be represented using $u$?

Vivian
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  • The answer uses that $A=A-0I$, then the zero eigenvectors are N(A-0I) (this by the definition of eigenvectors). You can see that since you have an eigenbase of different eigenvectors, $dim(N(A))=1$ (you can try to prove it), so all vectors are linear dependent to $u$. – user_sion Jun 06 '25 at 17:22

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Since $u,v,w$ are eigenvectors corresponding to distinct eigenvalues, they are linearly independent.

Hence, $u,v,w$ form a basis for $F^3$, where $F$ is the field you are working on.

Thus, if $x \in N(A)$, write $x=au+bv+cw$ for $a,b,c \in F$; then $$ 0 = Ax = aAu+bAv+cAw = bv+2cw $$ and this implies that $b=c=0$, so $x=au$.

azif00
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