MHB Proving $(T^2-I)(T-3I) = 0$ for Linear Operator $T$

[Guest2]

Problem: Let $T$ be the linear operator on $\mathbb{R}^3$ defined by

$$T(x_1, x_2, x_3)= (3x_1, x_1-x_2, 2x_1+x_2+x_3)$$

Is $T$ invertible? If so, find a rule for $T^{-1}$ like the one which defines $T$.

Prove that $(T^2-I)(T-3I) = 0.$


Attempt:

$(T|I)=\left[\begin{array}{ccc|ccc}
3 & 0 & 0 & 1 & 0 & 0 \\
0 & -1 &0 & 0 & 1 & 0 \\
2 & 1 & 1 & 0 & 0 & 1 \\
\end{array}\right] \to \left[\begin{array}{ccc|ccc}
1 & 0 & 0 & 1/3 & 0 & 0 \\
0 & 1 &0 & 0 & -1 & 0 \\
0 & 0 & 1 & -2/3 & 1 & 0 \\
\end{array}\right] = (I|T^{-1})$

Thus $T$ is invertible as it row-reduces to $I$ and

$$T^{-1} = \left[\begin{array}{ccc}
& 1/3 & 0 & 0 \\
& 0 & -1 & 0 \\
& -2/3 & 1 & 0 \\
\end{array}\right]$$

Thus the rule is $T^{-1}(x_1, x_2, x_3) = (\frac{1}{3}x_1, -x_2, -\frac{2}{3}x_1+x_2).$

But I'm stuck on showing that $(T^2-I)(T-3I) = 0$. I could directly manipulate this, but I think I'm supposed to somehow use linearity/earlier part of the question. Is this the case? Could someone please clarify.

 

[I like Serena]

Problem: Let $T$ be the linear operator on $\mathbb{R}^3$ defined by

$$T(x_1, x_2, x_3)= (3x_1, x_1-x_2, 2x_1+x_2+x_3)$$

Is $T$ invertible? If so, find a rule for $T^{-1}$ like the one which defines $T$.

Prove that $(T^2-I)(T-3I) = 0.$


Attempt:

$(T|I)=\left[\begin{array}{ccc|ccc}
3 & 0 & 0 & 1 & 0 & 0 \\
0 & -1 &0 & 0 & 1 & 0 \\
2 & 1 & 1 & 0 & 0 & 1 \\
\end{array}\right] \to \left[\begin{array}{ccc|ccc}
1 & 0 & 0 & 1/3 & 0 & 0 \\
0 & 1 &0 & 0 & -1 & 0 \\
0 & 0 & 1 & -2/3 & 1 & 0 \\
\end{array}\right] = (I|T^{-1})$

Thus $T$ is invertible as it row-reduces to $I$ and

$$T^{-1} = \left[\begin{array}{ccc}
& 1/3 & 0 & 0 \\
& 0 & -1 & 0 \\
& -2/3 & 1 & 0 \\
\end{array}\right]$$

Thus the rule is $T^{-1}(x_1, x_2, x_3) = (\frac{1}{3}x_1, -x_2, -\frac{2}{3}x_1+x_2).$

But I'm stuck on showing that $(T^2-I)(T-3I) = 0$. I could directly manipulate this, but I think I'm supposed to somehow use linearity/earlier part of the question. Is this the case? Could someone please clarify.

Hi Guest,

The matrix you've found for $T^{-1}$ is not actually the inverse.
In particular this shows because it's degenerate.
That is, it has a column with only zeroes.
More generally, its determinant is zero, meaning it's non-invertible.

The problem is that when you created $T$, the second row of the leftmost column should contain a $1$ instead of a $0$.

To show that $(T^2-I)(T-3I) = 0$, we might use the Caylay-Hamilton theorem, that says that the characteristic equation of a matrix also holds for the matrix itself.
Do you know what the characteristic equation is? (Wondering)

 

[Guest2]

Hi, I like Serena,

Thanks for the reply.

The characteristic equation is $P(\lambda) = \det(\lambda I - T) = (3-\lambda)(\lambda^2-1)$

So then $P(T) = 0 \implies (3I-T)(T^2-I) = 0 \implies (T^2-I)(T-3I) = 0.$

 

[Deveno]

To show $T$ is invertible, it suffices to show $T(v) = 0 \implies v = 0$.

This is not hard to demonstrate:

If $T(x_1,x_2,x_3) = (0,0,0)$, then $3x_1 = 0$, so $x_1 = 0$.

Now $T(0,x_2,x_3) = (0,-x_2,x_2+x_3)$, and if this is $(0,0,0)$ we must have $x_2 = 0$.

Finally, since $T(0,0,x_3) = (0,0,x_3)$ if the RHS is 0, we must have $x_3 = 0$.

Of course, this tells us $T^{-1}$ exists, but does not tell us what it IS.

But if $T(x_1,x_2,x_3) = (1,0,0)$, it follows that $(x_1,x_2,x_3) = T^{-1}(1,0,0)$, which would give us the first column of $T^{-1}$'s matrix (in the standard basis).

Equating $(3x_1,x_1-x_2,2x_1+x_2+x_3) = (1,0,0)$ gives us $x_1 = \frac{1}{3}$, which then tells us $x_2 = \frac{1}{3}$, and thus $x_3 = -1$.

Solving $T(x_1,x_2,x_3) = (0,1,0)$ and $T(x_1,x_2,x_3) = (0,0,1)$ similarly gives us the second and third columns of $T^{-1}$. You should then be able to give a similar form for $T^{-1}$ as was given for $T$.

I do not know if the Cayley-Hamilton theorem is available for you to use-that is the "lazy" way to go. Otherwise, you would have to compute the given product the "long way".

Knowing $T^3 - 3T^2 - T + 3I = 0$, does give us another way to compute $T^{-1}$:

$T^3 - 3T^2 - T = -3I$
$T(-\frac{1}{3}(T^2 - 3T - I)) = I$

so, evidently:

$T^{-1} = -\frac{1}{3}(T^2 - 2T - I)$

 

[Guest2]

...

This is such a nice method, thank you!