A question regarding the matrix matrix $P^{-1}DP$ where $D=diag(1,0,\ldots,0)$

Question

Let $A$ be an $n\times n$ matrix which is similar to $D=diag(1,0,\ldots,0)$ and Let $P$ be the invertible matrix such that $P^{-1}AP=D$. What is the vector $$ A \begin{pmatrix} 1\\1\\ \vdots \\1 \end{pmatrix}=? $$ My attempt: If the matrix is $2\times2$ and $$ P= \begin{pmatrix} a & b\\ c & d\\ \end{pmatrix} $$ where $$ \vec{v}= \begin{pmatrix} a\\ c \\ \end{pmatrix} $$ is the eigenvector of $1$, then we get $$ A \begin{pmatrix} 1\\1\\ \end{pmatrix}= P \begin{pmatrix} 1 & 0\\ 0 & 0\\ \end{pmatrix} P^{-1}\begin{pmatrix} 1\\1\\ \end{pmatrix}= \frac{d-b}{ad-bc}\begin{pmatrix} a\\c\\ \end{pmatrix} $$ How to generalized it to any order $n$?

Answer 1

If $P^{-1} A P = D = \operatorname{diag}(1,0,\ldots,0)$, then:

$$ A = P D P^{-1}$$

Note that the title presents this relationship differently than the body of your Question, but let's go with the latter. Also let's assume the matrix $A$ is meant to have real number entries.

Thus:

$$ A \begin{pmatrix} 1 \\ 1 \\ \vdots \\ 1 \end{pmatrix} = P D P^{-1} \begin{pmatrix} 1 \\ 1 \\ \vdots \\ 1 \end{pmatrix} $$

Without more information, not much more can be said about this vector. Indeed without further restrictions $A$ can be any rank $1$ matrix with trace $1$, i.e.

$$ A = u v^T $$

for some nonzero (column) vectors $u,v$ such that trace $A = \sum_{i=1}^n u_i v_i = 1$. Without loss of generality one can scale the vector $v$ to have entries that sum to $1$, by correspondingly multiplying the vector $u$ by the reciprocal of the sum of entries in $v$.

Now $v^T (1,1,\ldots,1)^T = r$, the sum of entries in $v$. It follows that:

$$ A \begin{pmatrix} 1 \\ 1 \\ \vdots \\ 1 \end{pmatrix} = r u $$

We can arrange for this vector to be any $u$ with Euclidean norm $1$, i.e. sum of entries squared is one, and scalar $r$ the sum of entries of $v$. Perhaps that is the conclusion you were meant to draw.

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The symmetric case is particularly easy to understand. Let $u$ be any (column) vector in $\mathbb R^n$ with Euclidean norm $1$.

Then $A = uu^T$ is a symmetric rank one matrix with trace $1$. Note that its diagonalized form is $D$ as described in this Question.

Now $A (1,1,\ldots,1)^T = u u^T (1,1,\ldots,1)^T = r u$ where scalar $r = \sum_{i=1}^n u_i$.