Question

Matrix $I_n$is similar to $2I_n$.

Short answer

The above statement is false.

Matrix$I_n$ is not similar to$2I_n$ , where is$I_n$ identity matrix of order n x n.

Step-by-step solution

Definition of similar matrix

A square matrix ‘B’ is said to be similar to a matrix A if there exists an invertible matrix P such that

$B=P^{-1}AP$

To prove matrix In is not similar to 2In

Let us take n = 2. Then,

$I_2=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right){\text{and2I}}_2=\left(\begin{array}{cc}2& 0\\ 0& 2\end{array}\right)$

Let us suppose that matrix $I_2$is similar to matrix$2I_2$ then there exists an invertible matrix $P=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$such that $2I_{2}P=P{I}_2.$

$$\begin{gathered} \Rightarrow \left(\begin{array}{cc}2& 0\\ 0& 2\end{array}\right)\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right) \\ \Rightarrow \left(\begin{array}{cc}2a& 2b\\ 2c& 2d\end{array}\right)=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right) \end{gathered}$$

$$\begin{gathered} \Rightarrow 2a=a \\ 2b=b \\ 2c=c \\ 2d=d \\ \Rightarrow a=0=b=c=d \end{gathered}$$

Thus, matrix$P=\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right)$ is not invertible and so there doesn’t exist any invertible matrix P such that matrix $I_2$is similar to matrix $2I_2$.

Since this holds for n = 2, therefore, it will hold for every value of n.

Hence, matrix $I_n$is not similar to $2I_n$ where is$I_n$identity matrix of order .

Final Answer

Since there doesn’t exist any invertible matrix P such that matrix $I_2$is similar to matrix $2I_2$for n = 2, therefore, there doesn’t exist any invertible matrix for any value of n.

Hence, matrix $I_n$is not similar to$2I_n$ where is$I_n$ identity matrix of order .