# Find all invariant subspace of $A$ [closed]

Consider a matrix mapping $A: V \to V$ for a vertor space V. Matrix $A$ has 3 eigenvalues are distinct : $\lambda_1,\lambda_2,\lambda_3$ and $v_1,v_2,v_3$ are vectors-corresponding .Find all invariant subspace of $A$

## closed as off-topic by Davide Giraudo, Grigory M, Alexander Gruber♦Jan 12 '15 at 17:58

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Hint: note that each $\text{span}(v_i)$ is a invariant vector space.

Also note that $\{0\}$ and $V$ are invariant subspaces. Furthermore note that if $A$ is an invariant subspace and $B$ is an invariant subspace then so is $A + B$.

• but it's not all. – Road Human Jan 12 '15 at 2:24
• @RoadHuman added some more hints. try to find all the invariant subspaces for each dimension. – Loreno Heer Jan 12 '15 at 2:34
• i see that, with dimension-1 has: $span(v_1),span(v_2),span(v_1+v_2),span(v_1+\sqrt{2}v_2),....$ it's countless ?? – Road Human Jan 12 '15 at 2:38
• @RoadHuman maybe this will help: math.stackexchange.com/questions/632701/… – Loreno Heer Jan 12 '15 at 2:40

The nontrivial invariant subspaces are just the eigenspaces, and sums of eigenspaces. Assuming this is a real vector space, these are $$U_1 = \{cv_1 : c\in\mathbb R\}, U_2 = \{cv_2 : c\in\mathbb R\}, U_3=\{cv_3 : c\in\mathbb R\}.$$ Observe that if $u\in U_j$ then $u=cv_j$ for some $c\in\mathbb R$, hence $Au = A(cv_j)=cAv_j=c\lambda v_j\in U_j$. This means that $U_j$ is invariant under $A$.

However, $U_1+U_2$, $U_1+U_3$, and $U_2+U_3$ are also invariant subspaces, as if $u_i\in U_i$ and $u_j\in U_j$, $i\ne j$ then $u_i = c_iv_i$ and $u_j=c_jv_j$ for some $c_i,c_j\in\mathbb R$. So $$A(u_i + u_j)= Au_i + Au_j = c_iAv_i + c_jAv_j = c_i\lambda_iv_i + c_j\lambda_jv_j\in U_i+U_j.$$

The question of finding all $A$-invariant subspaces is rather difficult in general. However it has a fairly easy answer in the case where $A$ is diagonalisable, as is the case here. The restriction of a diagonalisable linear operator to any invariant subspace is always diagonalisable, which implies that such a subspace is equal to the direct sum of its intersections with the eigenspaces of the operator. So the subspace is then precisely determined by specifying a subspace of each eigenspace. If in addition, as is the case here, all eigenspaces are of dimension$~1$, there are only two subspaces in each case: $\{0\}$ and the whole eigenspace. For $3$ such eigenspaces one therefore gets $2^3=8$ invariant subspaces.

In general there may well be infinitely many invariant subspaces, as happens for instance whenever an eigenspace has dimension more than$~1$. The most delicate case for classifying invariant subspaces is that of nilpotent operators (or those obtained by adding a scalar matrix to them).