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there is matrix $ B=\left(\begin{array}{ccc}1 & -1 \\ 1 & 3\\ \end{array}\right)$ ,

for vector $ v_1=\left(\begin{array}{ccc}1 \\ -1\\ \end{array}\right), B v_1=2v_1$ holds.
using $v_1$ and $v_2=\left(\begin{array}{ccc}a \\ b\\ \end{array}\right) $ which is a linear independent vector of magnitude 1, we can express $B.v_2$ as linear combination of $v_1$ & $v_2$ such as $B.v_2=\alpha v_1+2v_2$.
a.) find $v_2$ & $\alpha$

b.) $P=\left(\begin{array}{ccc}1 & a \\ -1 & b\\ \end{array}\right)$, then we can express $BP=PC$, find matrix C, and find $B^n$

first try to find $v_2$ & $\alpha$, $(B-2E)v_2=v_1$ $ B=[\left(\begin{array}{ccc}1 & -1 \\ 1 & 3\\ \end{array}\right)$ - $ \left(\begin{array}{ccc}2 & 0 \\ 0 & 2\\ \end{array}\right) ] .v_2 =\left(\begin{array}{ccc}1 \\ -1\\ \end{array}\right)$
got $v_2=\left(\begin{array}{ccc}-b-1 \\ b \end{array}\right)$ = $\left(\begin{array}{ccc}-1 \\ 1\\ \end{array}\right)s + \left(\begin{array}{ccc}-1 \\ 0\\ \end{array}\right)t$, since $v_2$ is vector with magnitude 1 , i chose $ \left(\begin{array}{ccc}-1 \\ 0\\ \end{array}\right)$ with $\alpha =1$, is my answer right?
then find matrix $C$ , i know matrix $C$ is matrix diagonal, $P^{-1}BP=C$, so with $P=\left(\begin{array}{ccc}1 & -1\\ -1 & 0\\ \end{array}\right)$ , after some calculation $ C=\left(\begin{array}{ccc}2 & 0 \\ 1 & 2\\ \end{array}\right)$

But i wasnt so sure to find $B^n$ ?

José Carlos Santos
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fiksx
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2 Answers2

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Credit: José Carlos Santos for pointing out that $C= \begin{bmatrix} 2 & 1 \\ 0 & 2\end{bmatrix}$

Guide:

$$P^{-1}BP=C$$

$$P^{-1}B^nP=C^n$$

$$C=\begin{bmatrix} 2 & 1 \\ 0 & 2 \end{bmatrix}$$

$$C^2 = \begin{bmatrix} 4 & 4 \\ 0 & 4 \end{bmatrix}=\begin{bmatrix} 2^2 & 2(2) \\ 0 & 2^2 \end{bmatrix}$$

$$C^3 = \begin{bmatrix} 8 & 0 \\ 12 & 8 \end{bmatrix}=\begin{bmatrix} 2^3 & 3(2^2) \\ 0 & 2^3 \end{bmatrix}$$

Compute a few more terms, spot a pattern and prove the pattern of $C^n$ perhaps by induction.

Edit:

$\begin{bmatrix} a \\ b\end{bmatrix}= \begin{bmatrix} -1 \\ 1\end{bmatrix} b + \begin{bmatrix} -1 \\ 0\end{bmatrix}$

Note that there is no $t$ at the end.

Since you were solving $$(B-2I)v_2 = v_1$$

You have chosen $\alpha = 1$.

If you want to impose the condition that $v_2$ is of length $1$.

You just have to solve for $b$ in $\left\|\begin{bmatrix} -1 \\ 1\end{bmatrix} b + \begin{bmatrix} -1 \\ 0\end{bmatrix}\right\|^2=1$, no guessing is required.

Siong Thye Goh
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  • thanks!!, but i saw some problem of computing power of matrix, with this relationship $P^{-1}B^nP=C^n$ , then $C=\left(\begin{array}{ccc}2^n & 0 \ 0 & 2^n\ \end{array}\right)$ , why it not applicable here? and also suppose it is big matrix like 3X3 then to compute $C^n$ will be hard, and hard to find the pattern too. so is there another way(?) – fiksx Dec 08 '17 at 10:24
  • It is not applicable because $C$ is not a diagonal matrix for this particular problem. We know the general format for the power of Jordan block. For example, here – Siong Thye Goh Dec 08 '17 at 10:29
  • C is not diagonal matrix, because B rank is 1 but the geometry multiplicity is two for $\lambda =2$? so it is not diagonal matrix? since it is diagonal matrix if gm=am, but here gm<am, so i have to use jordan block? but is this matrix diagonalizable(?) – fiksx Dec 08 '17 at 11:48
  • rank of $B$ is $2$. the question is what is the nullity of $B-\lambda I$. It is $1$ in this case. The geometric multiplicity is $1$, but the algebraic multiplicity is $2$, hence it is not diagonalizable. Every matrix has a jordan canonical form, which is more general than diagonalization. I am guessing you might be asked this question on a diagonalization question to test if you can figure out the relation of $P^{-1}B^nP=C^n$. – Siong Thye Goh Dec 08 '17 at 12:07
  • yes, i mean rank P is 1 and the nullity is 1 same as geometry multiplicity , but because algebraic multiplicity is 2, so it has 1 eigen vector only, it is not diagonalizable, but suppose the nullity of $B-\lambda I=4$ and rank= 0 (is it possible?), then it is diagonalizable with $\lambda $ in the diagonal and all $0$ , is my assumption true? yes, i never heard jordan block before. – fiksx Dec 08 '17 at 12:51
  • rank and nullity are both nonnegative and them sum up to the number of columns, hence it is impossible. If the nullity of $B-\lambda I$ is $2$, then the geometric multiplicity for eigenvalue $\lambda$ is $2$. In that case, then it is diagonalizable. The diagonal entries will be $\lambda$. It is just a name. You will learn it eventually if you major in maths. – Siong Thye Goh Dec 08 '17 at 12:58
  • okay!! thankyou so much for the clear explanation, one more question, $\left(\begin{array}{ccc} -1 & -1 \ 1 & 1\ \end{array}\right) .v_2 =\left(\begin{array}{ccc}1 \ -1\ \end{array}\right)$ then $v_2=\left(\begin{array}{ccc}-b-1 \ b \end{array}\right)\left(\begin{array}{ccc}-2 \ 1\ \end{array}\right)$ ,am i right? but this also true $\left(\begin{array}{ccc}-1 \ 0\ \end{array}\right)$, and the question want vector that has magnitude for $v_2$ , should i use trial and error to find $v_2$? – fiksx Dec 08 '17 at 16:11
  • Responded in my edit. No trial and error is required. – Siong Thye Goh Dec 08 '17 at 21:14
  • thankyou so much!! so b will be $b=0$ or $ b=-1$ and the vector is $\left(\begin{array}{ccc}-1 \ 0\ \end{array}\right)$ or $\left(\begin{array}{ccc}0 \ -1\ \end{array}\right)$ . thanks for really clear explanation!! – fiksx Dec 10 '17 at 05:18
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It looks like this exercise is taking you step-by-step through calculating what’s known as the Jordan normal form of $B$ (in this exercise, the matrix $C$) and then using that to compute $B^n=(PCP^{-1})^n=PC^nP^{-1}$. It’s not too hard to work out the pattern for $C^n$, as illustrated in other answers, but you can save yourself some guesswork by decomposing $C$ into the sum of the diagonal matrix $2I$ and a nilpotent matrix $N$, i.e., $$C=\begin{bmatrix}2&1\\0&2\end{bmatrix} = \begin{bmatrix}2&0\\0&2\end{bmatrix}+\begin{bmatrix}0&1\\0&0\end{bmatrix} = 2I+N.$$ Now, use the Binomial theorem and the fact that $2I$ and $N$ commute to expand $$(2I+N)^n = 2^nI+n2^{n-1}N+\binom n 2 2^{n-2}N^2 + \cdots$$ but $N^2=0$, as you can verify for yourself, therefore $$C^n = 2^nI+n2^{n-1}N = \begin{bmatrix}2^n & n2^{n-1} \\ 0 & 2^n\end{bmatrix}.$$

In fact, you can use a similar decomposition to compute $B^n$ more directly, without finding the matrix $P$ first. We have $$B = PCP^{-1} = P(2I+N)P^{-1} = 2PIP^{-1}+PNP^{-1} = 2I+PNP^{-1}.$$ Now, $$(PNP^{-1})^2 = PNP^{-1}PNP^{-1} = PN^2P^{-1} = 0,$$ so by a similar computation to that above we get $$B^n = 2^nI + n2^{n-1}(B-2I) = \begin{bmatrix}2^n&0\\0&2^n\end{bmatrix}+\begin{bmatrix}-n2^{n-1}&-n2^{n-1}\\n2^{n-1}&n2^{n-1}\end{bmatrix} = \begin{bmatrix}2^n-n2^{n-1} & -n2^{n-1} \\ n2^{n-1} & 2^n+n2^{n-1}\end{bmatrix}.$$ This is a lot less work than working out generalized eigenvectors of $B$ (the matrix $P$ in this exercise) and then doing all of those matrix multiplications. You can use this method to compute powers and exponentials of any $2\times2$ matrix with a repeated eigenvalue that isn’t a multiple of the identity (and the method can be generalized to some extent to larger matrices).

amd
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  • wow this method really effective!! $(PNP^{-1})^2 = PNP^{-1}PNP^{-1} = PN^2P^{-1} = 0,$, is this mean that $B^2=0$ and so on will be 0 right? is this method only applicable to matrix with repeated eigen value in diagonal? and not applicable to for example $(\lambda -2)^2 ( \lambda-1)$ ? – fiksx Dec 10 '17 at 06:20
  • @Vixf No, it’s the square and higher powers of $(B-\lambda I)$ that vanish. The method depends on the fact that $\lambda I$ commutes with everything so that you can rearrange the products to bring powers of $N$ together. A general diagonal matrix doesn’t have this property, so this doesn’t work for mixed eigenvalues. There are other short-cuts that you can use for other cases, but at some point it’s just easier to compute the full Jordan decomposition. – amd Dec 10 '17 at 18:18
  • ok thankyou so much for the clear explanation!!!! – fiksx Dec 13 '17 at 11:32