Suppose $A$ is any $3×3$ non-singular matrix and $(A−3I)(A−5I)=O$, where $I=I_3$ and $O=O_3$. If $αA+βA^{−1}=4I$, then $α+β$ is equal to ___ .

This seems like a simple and straightforward question, I expanded out the given condition,

$$(A−3I)(A−5I)=O \\\implies A^2 -8A +15I = O $$ and then pre-multipied $A^{-1}$ to both sides which gives: $$A^{-1} \cdot (A^2 -8A +15I )= A^{-1} \cdot O \\\implies A - 8I +15A^{-1} = O \\\text{or}\ \frac{1}2A +\frac{15}{2}A^{-1} = 4I$$ Comparing with what was given in the question,$ \alpha = \frac{1}2$ and $ \beta = \frac{15}2$ which gives the sum to be 8.

But then I thought, how is this possible? From the equation $(A−3I)(A−5I)=O$, it seems $A$ can be either $3I$ or $5I$ (definitely not both at the same time), when I substiuted $ A = 3I$ in $αA+βA^{−1}=4I$, I get both α and β in the same equation, if the other possibility is substituted, we get another equation which gives the same result as above when the two are solved simultaneously. But this can't be possible? What is actually the value of $A$ then, since I'm getting such a contradiction?

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    $\begingroup$ think about characteristic roots of $A$....one more thing If $AB=O$ then either $A$ is null or $B$ is null or both will be singular.. $\endgroup$ – освящение May 2 '18 at 15:56

You are thinking as if matrices were numbers. The product $(A-3I)(A-5I)=0$ does not tell you that one of the factors is zero; it only tells you that their product is zero.

Matrices that satisfy your polynomial are, among others, $$ A_1=\begin{bmatrix} 3&0&0\\0&3&0\\ 0&0&5\end{bmatrix}, \ \ A_2=\begin{bmatrix} 5&0&0\\0&5&0\\0&0&3\end{bmatrix}. $$ If you take $S$ to be any invertible matrix, $SA_1S^{-1}$ and $SA_2S^{-1}$ will also satisfty the equation. If $p(t)=(t-3)(t-5)$ is the minimal polynomial for $A$, then the above are all the possible options.

When you conclude from $\frac12\,A+\frac{15}2\,A^{-1}=4I$ that $\alpha=\frac12$, $\beta=\frac{15}2$, you are using that $A$ and $A^{-1}$ are linearly independent, which they are in the above case.

But $A=3I$ and $A=5I$ also satisfy the equation $(A-3I)(A-5I)=0$, but now $A$ and $A^{-1}$ are not linearly independent. In the first case, from $$ 4I=\alpha A+\beta A^{-1}=(3\alpha+\frac{\beta}3)\,I, $$ we get $9\alpha+\beta=12$. This gives us infinitely many choices of $\alpha,\beta$. For instance, if $\alpha=1$, then $\beta=3$ and $\alpha+\beta=4$; if $\alpha=2$, then $\beta=-6$ and $\alpha+\beta=-4$; if $\alpha=1/3$ then $\beta=9$, and $\alpha+\beta=28/3$.

For the case $A=5I$, we also get infinitely many choices.


Two cases:

  • If $A-3I=0$, you obtain the relation $$3\alpha I+\frac \beta3I=4I\iff9\alpha+\beta=12,$$ so you can't say whatever about $\alpha+\beta$. Similarly if $A=5I$.
  • If $A-3I\ne 0\ne A-5I$, then $t^2-8t+15$ is the minimal polynomial of $A$, and the relation between $A$ and $A^{-1}$ can be re-written as \begin{align} &&&\alpha A^2-4A+\beta=0\implies\frac\alpha1=\frac{-4}{-8}=\frac\beta{15}=\frac{\alpha+\beta}{1+15}\\ &\text{so }&& \frac{\alpha+\beta}{16}=\frac12\iff\alpha+\beta=8. \end{align}


$$ A^2-8A+15I=O\Rightarrow A^2=8A-15I $$

then from

$$ \alpha A^2-4A+\beta I = O \Rightarrow A = \frac{\beta-15\alpha}{8\alpha-4}I = \mu I $$

Then equivalently we have

$$ \mu^2-8\mu+15=0\\ \alpha\mu^2-4\mu+\beta = 0 $$


$$ \mu - 5= 0\\ \mu - 3= 0\\ \alpha\mu^2-4\mu+\beta = 0 $$


$$ 25\alpha-20 +\beta = 0\\ 9\alpha -12+\beta = 0 $$

and solving we get

$$ \alpha+\beta = 8 $$


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