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Good morning,

How does one find the subspaces that are invariant under $A$ for $$A = \begin{pmatrix} 1 & 0 & 0 & 0\\ 0 &2 & 0 & 0\\ 0 & 0 & 2 & 0\\ 0 & 0 & 0 & 3 \end{pmatrix}\ \in M_{4} (\mathbb{R}).$$

Thank you.

Arturo Magidin
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  • Is that supposed to be a diagonal matrix? – Arturo Magidin Apr 05 '11 at 19:39
  • Can you name any of the invariant subspaces of that matrix? – Jonas Meyer Apr 05 '11 at 19:39
  • @Arturo: Yeah, what was it before your edit? –  Apr 05 '11 at 19:46
  • @Nir: My edit only fixed the misspelling "spesific" to "specific". You can see what changes were made by clicking on the "xxx mins ago" link next to "edit." So... please don't be lazy and put in those missing 0s. – Arturo Magidin Apr 05 '11 at 19:47
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    @Nir: For starters, can you name the invariant subspaces of the identity matrix? – Jonas Meyer Apr 05 '11 at 19:47
  • @Arturo: Put in the what? –  Apr 05 '11 at 19:49
  • @ jonas: ok, so for start I know there are three eigenvalues and three eigenspaces, and what exactly do I need to do here? –  Apr 05 '11 at 19:50
  • @Nir: Sigh. The missing 0s. Clearly, something others will have to do for you, as seems to be often the case. You really should spend some effort making your posts readable. – Arturo Magidin Apr 05 '11 at 19:50
  • @Arturo: oh the zeros. –  Apr 05 '11 at 19:53
  • @Arturo: sorry I thought it is obvious that there are zeros. the way you write your comment makes me feel uncomfterbal. –  Apr 05 '11 at 19:57
  • you have a basis of eigenvectors (ie one dimensional invariant subspaces) choosing any two of those gives you invariant planes, any three gives you an invariant 3-subspace – yoyo Apr 05 '11 at 20:26
  • @Nir: Here's something to get you started. If $v$ is an eigenvector of $A$, then the span of $v$, i.e. $\mathbb{R}v={rv:r\in\mathbb R}$ is an invariant subspace of $A$. Similarly, any eigenspace is invariant, and any subspace of an eigenspace is invariant. (The reason I asked about the identity is because on each eigenspace, $A$ acts as a scalar multiple of the identity matrix.) Perhaps you can get further in terms of describing all invariant subspaces of $A$? – Jonas Meyer Apr 05 '11 at 20:28

2 Answers2

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A subspace $\mathbf{W}$ is invariant under $A$ if and only for every $\mathbf{w}\in\mathbf{W}$, $A\mathbf{w}\in\mathbf{W}$.

The only $0$-dimensional subspace is always invariant (for any matrix), since $A\mathbf{0}=\mathbf{0}\in\{\mathbf{0}\}$.

$\mathbf{V}$ itself (in this case, $\mathbb{R}^4$) is also always invariant, since $A\mathbf{v}\in\mathbb{R}^4$ for every $\mathbf{v}\in\mathbb{R}^4$.

So, let's deal with the in-betweens:

  1. A $1$-dimensional invariant subspace; well, a $1$-dimensional subspace is of the form $\mathbf{W} = \{\alpha\mathbf{w}_0\mid\alpha\in\mathbb{R}\}$ for some fixed nonzero vector $\mathbf{w}_0$. If such a $\mathbf{W}$ is invariant, then $A(\alpha\mathbf{w}_0)=\alpha A\mathbf{w}_0\in\mathbf{W}$ for every scalar $\alpha$. This happens if and only if $A\mathbf{w}_0\in\mathbf{W}$ (prove the equivalence). What does that tell you about $\mathbf{w}_0$?

  2. A $2$-dimensional invariant subspace. This would be of the form $\mathbf{W}=\{\alpha\mathbf{w}_1 +\beta\mathbf{w}_2\mid\alpha,\beta\in\mathbb{R}\}$. Such a subspace is invariant if and only if $A\mathbf{w}_1\in\mathbf{W}$ and $A\mathbf{w}_2\in\mathbf{W}_2$ (prove it). There are several ways this can happen in principle, but here, since $A$ is inverible, then $A\mathbf{w}_1$ and $A\mathbf{w}_2$ would have to span $\mathbf{W}$ itself, which reduces the possibilities somewhat.

  3. Analogous with $3$-dimensional invariant subspaces.

Here, $A$ is diagonalizable (in fact, diagonal). You know that $A(a,b,c,d) = (a,2b,2c,3d)$. This makes it pretty straightforward to check when you get an invariant subspace by writing these hypothetical bases for the invariant subspaces, and seeing what it means for their images to lie in the space they span.

Or even better, you can use an important theorem:

Theorem. Let $\mathbf{T}\colon\mathbf{V}\to\mathbf{V}$ be a linear operator on a finite dimensional vector space. If $\mathbf{T}$ is diagonalizable, and $\mathbf{W}$ is a $\mathbf{T}$-invariant subspace of $\mathbf{W}$, then the restriction of $\mathbf{T}$ to $\mathbf{W}$, $\mathbf{T}_{\mathbf{W}}$, is also diagonalizable.

This will tell you what the $2$- and $3$-dimensional invariant subspaces look like in terms of the $1$-dimensional subspaces.

(There are many proofs of the theorem; the least computational I know uses the facts that $\mathbf{T}$ is diagonalizable if and only its characteristic polynomial splits and the minimal polynomial is squarefree; and that the characteristic and minimal polynomials of the restriction of $\mathbf{T}$ to an invariant subspace divide the characteristic and minimal polynomials of $\mathbf{T}$, respectively).

Arturo Magidin
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4

Define $\mathscr{T}:\mathbb{R}^{4}\rightarrow\mathbb{R}^{4}$ by $\mathscr{T}(\mathbf{x})=A\mathbf{x}$ for $\mathbf{x}\in \mathbb{R}^{4},$ where $$A=\begin{pmatrix} 1& 0& 0& 0\\ 0& 2& 0& 0\\ 0& 0& 2& 0\\ 0& 0& 0& 3 \end{pmatrix}\in M_{4} (\mathbb{R}).$$ $$\mathbf{e}_{1}=(1,0,0,0)^{T},\mathbf{e}_{2}=(0,1,0,0)^{T},\mathbf{e}_{3}=(0,0,1,0)^{T},\mathbf{e}_{4}=(0,0,0,1)^{T}.$$

The complete list of $\mathscr{T}$-invariant subspaces :

  • Zero-dimensional subspace: $\{\mathbf{0}\}.$
  • One-dimensional subspaces:$$\text{Span}\lbrace \mathbf{e}_{1}\rbrace,\text{Span}\lbrace \mathbf{e}_{2}\rbrace,\text{Span}\lbrace \mathbf{e}_{3}\rbrace, \text{Span}\lbrace \mathbf{e}_{4}\rbrace,\text{Span}\lbrace \mathbf{e}_{2}+k\mathbf{e}_{3}\rbrace(k\ne 0).$$

  • Two-dimensional subspaces: $$\text{Span}\lbrace \mathbf{e}_{1},\mathbf{e}_{2}\rbrace,\text{Span}\lbrace \mathbf{e}_{1},\mathbf{e}_{3}\rbrace,\text{Span}\lbrace\mathbf{e}_{1},\mathbf{e}_{4}\rbrace, \text{Span}\lbrace\mathbf{e}_{2},\mathbf{e}_{3}\rbrace,\text{Span}\lbrace \mathbf{e}_{2},\mathbf{e}_{4}\rbrace,\text{Span}\lbrace\mathbf{e}_{3},\mathbf{e}_{4}\rbrace,\text{Span}\lbrace\mathbf{e}_{1},\mathbf{e}_{2}+k\mathbf{e}_{3}\rbrace(k\ne 0),\text{Span}\lbrace\mathbf{e}_{4},\mathbf{e}_{2}+k\mathbf{e}_{3}\rbrace(k\ne 0). $$

  • Three-dimensional subspaces:

$$\text{Span}\lbrace \mathbf{e}_{1},\mathbf{e}_{2},\mathbf{e}_{3}\rbrace,\text{Span}\lbrace \mathbf{e}_{2},\mathbf{e}_{3},\mathbf{e}_{4}\rbrace,\text{Span}\lbrace \mathbf{e}_{1},\mathbf{e}_{2},\mathbf{e}_{4}\rbrace,\text{Span}\lbrace \mathbf{e}_{1},\mathbf{e}_{3},\mathbf{e}_{4}\rbrace,\text{Span}\lbrace \mathbf{e}_{1},\mathbf{e}_{4},\mathbf{e}_{2}+k\mathbf{e}_{3}\rbrace(k\ne0).$$

  • Four-dimensional subspace:$\text{Span}\lbrace\mathbf{e}_{1},\mathbf{e}_{2},\mathbf{e}_{3},\mathbf{e}_{4}\rbrace.$
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    Why the 1- dimensional subspaces contain span { e_2 + k e_3} (k not equal 0) .... I can not see that it is a 1-dimensional space .... could you explain this for me please? – Emptymind Jan 28 '19 at 12:45