Finding Eigenspaces of a Matrix

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\begin{bmatrix}<br /> 3 &amp; 2\\<br /> 4 &amp; 1<br /> \end{bmatrix}
det(A-\lambda I)=\begin{vmatrix}<br /> 3-\lambda &amp; 2\\<br /> 4 &amp; 1-\lambda<br /> \end{vmatrix}=(3-\lambda)(1-\lambda)-8=\lambda^2-4\lambda-5
\lambda_{1}=5 and \lambda_{2}=-1
When \lambda=5, \begin{bmatrix}<br /> -2 &amp; 2\\<br /> 4 &amp; -4<br /> \end{bmatrix}\Rightarrow \begin{bmatrix}<br /> 1 &amp; -1\\<br /> 0 &amp; 0<br /> \end{bmatrix}
The eigenspace for \lambda_{1} is \begin{bmatrix}<br /> 1\\<br /> 1<br /> \end{bmatrix}
When \lambda=-1, \begin{bmatrix}<br /> 4 &amp; 2\\<br /> 4 &amp; 2<br /> \end{bmatrix}\Rightarrow \begin{bmatrix}<br /> 1 &amp; 0\\<br /> 0 &amp; 1<br /> \end{bmatrix}
The eigenspace for\lambda_{2} is \begin{bmatrix}<br /> 0\\<br /> 0<br /> \end{bmatrix}

I don't know what is going wrong but my second Eigenspace is wrong compared to the books answer which is \begin{bmatrix}<br /> 1\\<br /> -2<br /> \end{bmatrix}
 
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Dustinsfl said:
\begin{bmatrix}<br /> 3 &amp; 2\\<br /> 4 &amp; 1<br /> \end{bmatrix}
det(A-\lambda I)=\begin{vmatrix}<br /> 3-\lambda &amp; 2\\<br /> 4 &amp; 1-\lambda<br /> \end{vmatrix}=(3-\lambda)(1-\lambda)-8=\lambda^2-4\lambda-5
\lambda_{1}=5 and \lambda_{2}=-1
When \lambda=5, \begin{bmatrix}<br /> -2 &amp; 2\\<br /> 4 &amp; -4<br /> \end{bmatrix}\Rightarrow \begin{bmatrix}<br /> 1 &amp; -1\\<br /> 0 &amp; 0<br /> \end{bmatrix}
The eigenspace for \lambda_{1} is \begin{bmatrix}<br /> 1\\<br /> 1<br /> \end{bmatrix}
When \lambda=-1, \begin{bmatrix}<br /> 4 &amp; 2\\<br /> 4 &amp; 2<br /> \end{bmatrix}\Rightarrow \begin{bmatrix}<br /> 1 &amp; 0\\<br /> 0 &amp; 1<br /> \end{bmatrix}
Your mistake is above. The [4 2; 4 2] matrix doesn't row reduce to the identity matrix. Try again.
Each matrix for calculating the eigenspace can't reduce to the identity; otherwise its determinant would not be zero.
Dustinsfl said:
The eigenspace for\lambda_{2} is \begin{bmatrix}<br /> 0\\<br /> 0<br /> \end{bmatrix}

I don't know what is going wrong but my second Eigenspace is wrong compared to the books answer which is \begin{bmatrix}<br /> 1\\<br /> -2<br /> \end{bmatrix}
 


I had a -2 entered into my calc.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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