We guess that a $3\times 3$ matrix $A$ has eigenvalues $0, 3, 5$ with eigenvectors $u, v, w$. a) Find a base of $N(A)$, and a base of $R(A)$. b) Find a solution of the equation $Ax = v + w$. Find all the solutions. c) Show that $Ax= u$ has no solutions.
I don't really know what to do. That's all I thought: $tr(A) = 8$, $det(A)= 0$ and: we have a $3\times 3$ matrix with $3$ different eigenvalues, so $u, v, w$ are linearly independent and $A$ is diagonalizable.
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Since the eigenvalues are distinct, the eigenvectors span the entire space.
Hence any $x$ can be written uniquely as $x=x_1u +x_2 v +x_3 w$.
Since $Ax = 3x_2 v +5x_3 w$, we see that (i) $Ax = 0$ iff $x_2=x_3 = 0$ iff $x \in \operatorname{sp}\{u\}$, and (ii) $y \in {\cal R} A$ iff $y \in \operatorname{sp}\{v,w\}$.
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Let’s start with the first part of the question. $N(A)$ is the set of vectors $x$ for which $Ax=0$. This equation is equivalent to $Ax=0x$. Does that look familiar?
For the range of $A$, we have three distinct eigenvalues, so the three given eigenvectors form a basis for $\mathbb R^3$. By linearity, $A(au+bv+cw)=aAu+bAv+cAw$. Now apply the fact that these are eigenvectors to eliminate $A$ from the right-hand side. What you will end up with tells you that every element of $R(A)$ is a linear combination of a certain set of known vectors. For these to be a basis, you will also need to show that those vectors are linearly independent.
Once you’ve worked out a basis for $R(A)$, the answer to part (b) should be fairly obvious. For part (c), recall that for a solution to exist, we must have $u\in R(A)$ and use your result for the basis of $R(A)$ from the first part.
Since $u,v$, and $w$ are linearly independent, you can write any relevant vector as $c_1u + c_2v + c_3w$.
Now, for a), you want to find $x$ such that $Ax = 0$, so think about at $A(c_1u + c_2v + c_3w)$ = 0.
For b), what kinds of vectors $b$ do you get for $A(c_1 u + c_2v + c_3w) = b$?
For c), think about $A(c_1 u + c_2v + c_3w) = u$