Practise: Directions That Stay Put

The characteristic polynomial is ${\lambda }^2-7\lambda +6$. Its discriminant is $49-24=25$, whose square root is $5$, so the roots are $(7+5)/2=6$ and $(7-5)/2=1$. Both are positive, so the matrix stretches two directions and flips neither. The two checks from section 9.1 hold: the sum $6+1=7$ equals the trace, and the product $6\cdot 1=6$ equals the determinant. The directions can be recovered if you want them. For $\lambda =6$, the equation $(A-6I)v=0$ reads $-v_1+2v_2=0$, so $v=(2,1)$ works, and $A(2,1)=(12,6)=6(2,1)$. For $\lambda =1$, the second row of $A-I$ reads $2v_1+v_2=0$, so $v=(1,-2)$ works, and $A(1,-2)=(1,-2)$. The polynomial alone does not hand you the directions; it hands you the stretch factors, and each one costs one more small solve.

Since $A(1,1)=(5,5)=5(1,1)$, the vector $(1,1)$ is an eigenvector with eigenvalue $5$. The sum of the eigenvalues equals the trace, which is $4+3=7$. The second eigenvalue is therefore $7-5=2$. Two independent checks confirm it. The product $5\cdot 2=10$ equals the determinant $4\cdot 3-1\cdot 2=10$. And the eigenvector for $\lambda =2$ is not $(1,-1)$; solving $(A-2I)v=0$ with $A-2I=\left(\begin{array}{cc}2& 1\\ 2& 1\end{array}\right)$ gives $2v_1+v_2=0$, so $v=(1,-2)$, and $A(1,-2)=(2,-4)=2(1,-2)$. Reading one eigenvalue from a direction you were handed is the cheap half of the problem. The trace supplies the other eigenvalue, and no polynomial is needed.

After $t$ steps the state is $c_1{\lambda }_1^t{v}_1+c_2{\lambda }_2^t{v}_2$. With $c_1=1$, $c_2=2$, ${\lambda }_1=3$, ${\lambda }_2=1$ and $t=4$, the first term has size $1\cdot 3^4=81$ and the second has size $2\cdot 1^4=2$. The ratio is $2/81=0.02469\dots$, which is $0.025$ to three decimal places. The same ratio at step $t$ is $2/3^t$, because the eigenvalue $1$ leaves its coefficient alone while the eigenvalue $3$ triples the other one. So the second term falls by a factor of $3$ at every step. That shrinking ratio, not the size of either term on its own, is what makes the dominant direction win. Here the domination is quick: after ten steps the ratio is $2/59049$, about $0.000034$.

The matrix is triangular, so its eigenvalues are the diagonal entries $3$ and $2$. Check the directions: $A(1,0)=(3,0)=3(1,0)$, and $A(1,-1)=(2,-2)=2(1,-1)$. The two eigenvectors are independent, so they form an eigenbasis. Write the start in that basis: $(1,1)=2(1,0)-1(1,-1)$, because $2(1,0)-(1,-1)=(2-1,0+1)=(1,1)$. Then $$A^t(1,1)=2\cdot 3^t(1,0)-1\cdot 2^t(1,-1)=\left(2\cdot 3^t-2^t,2^t\right).$$ At $t=2$ this gives $(18-4,4)=(14,4)$. The direct check runs $A(1,1)=(3+1,2)=(4,2)$ and then $A(4,2)=(12+2,4)=(14,4)$, the same pair. The formula also predicts step three at $(54-8,8)=(46,8)$, and $A(14,4)=(42+4,8)=(46,8)$. The two routes agree because the eigenbasis turns one matrix multiplication into two scalar multiplications. Nothing in the formula is a matrix multiplication; the off-diagonal $1$ only fixed the eigenvector $(1,-1)$, and that eigenvector fixed the coefficients $2$ and $-1$.

After $n$ steps the contamination has been multiplied by $(4/5{)}^n$. You want $(0.8{)}^n\le 0.001$. Taking logarithms gives $n\ln (0.8)\le \ln (0.001)$. Both logarithms are negative, so dividing flips the inequality: $n\ge 30.954$. The count must be a whole number, so $31$ steps. The two sides sit close together: $0.8^{30}=0.001238$, which is still above one thousandth, and $0.8^{31}=0.00099$, which is below it. Thirty steps would miss the target by about a quarter, so the rounding matters here. The lesson is the one section 9.3 draws from its $2$ and $1.9$ pair: a small gap is expensive, and the cost grows like the logarithm of the requested accuracy divided by the logarithm of the ratio. Moving the ratio from $0.8$ to $0.9$ raises the count from $31$ to $66$.

First find the eigenvalues. $A(1,1)=(5+3,3+5)=(8,8)=8(1,1)$, so $(1,1)$ has eigenvalue $8$. And $A(1,-1)=(5-3,3-5)=(2,-2)=2(1,-1)$, so $(1,-1)$ has eigenvalue $2$. Their dot product is $1-1=0$, so after dividing by $\sqrt{2}$ the two unit vectors are perpendicular. The matrix $Q$ has those unit vectors as columns. Multiplying $A$ on the right by $Q$ sends each column to its eigenvalue times itself, and multiplying on the left by $Q^{\top }$ reads off the coordinates in the eigenbasis. The result is $$Q^{\top }AQ=\left(\begin{array}{cc}8& 0\\ 0& 2\end{array}\right).$$ The zero in the top-right corner is the dot product $(1,1)\cdot (1,-1)/2=0$: it vanishes because the eigenvectors are perpendicular. That is the spectral theorem in one calculation. The two stretch factors are $8$ and $2$, so the map amplifies the diagonal four times as much as the anti-diagonal.

Compute trace and determinant for each. For (a), the trace is $8$ and the determinant is $16-1=15$; the eigenvalues solve ${\lambda }^2-8\lambda +15=0$, which factors as $(\lambda -5)(\lambda -3)$, so both are $5$ and $3$ and the form is a bowl. For (b), the determinant is $-8$, so the eigenvalues have opposite signs and the form is a saddle. For (c), the determinant is $0$ and the trace is $2$, so the eigenvalues are $2$ and $0$; the form vanishes at $(1,-1)$, because $(1,-1{)}^{\top }A(1,-1)=0$, and it is positive elsewhere. That is a trough, not a bowl: the condition $x^{\top }Ax>0$ fails at a nonzero vector. For (d), the determinant is $3$ and the trace is $-4$; the eigenvalues solve ${\lambda }^2+4\lambda +3=0$, which factors as $(\lambda +1)(\lambda +3)$, so both are negative and the form curves down in every direction. The test in section 9.5 uses the signs of the eigenvalues. The determinant and trace deliver those signs without solving the polynomial: positive determinant means the two eigenvalues share a sign, and the trace then says which sign. Only (a) passes both checks.

Multiplying $A$ by itself gives $A^{\top }A=A^2=\left(\begin{array}{cc}1+4& 2+2\\ 2+2& 4+1\end{array}\right)=\left(\begin{array}{cc}5& 4\\ 4& 5\end{array}\right)$. Its trace is $10$ and its determinant is $25-16=9$, so its eigenvalues solve ${\lambda }^2-10\lambda +9=0$, which factors as $(\lambda -9)(\lambda -1)$. The eigenvalues are $9$ and $1$, and the singular values are their square roots, $3$ and $1$. Two checks hold. The product $3\cdot 1=3$ equals $|\det A|=|1-4|=3$. And the right singular vectors are the eigenvectors $(1,1)/\sqrt{2}$ and $(1,-1)/\sqrt{2}$ of $A^{\top }A$; sending the first through $A$ gives $(3,3)/\sqrt{2}$, of length $3$, and the second gives $(-1,1)/\sqrt{2}$, of length $1$. Both singular values are nonzero, so the rank is $2$: the map sends the unit circle to an ellipse with semi-axes $3$ and $1$, not to a line segment.