Question

If matrix A is similar to matrix B, and matrix B is similar to matrix C, then C must be similar to matrix A.

Short answer

The above statement is false.

If matrix A is similar to matrix B, and matrix B is similar to matrix C, then there does not exist any invertible matrix S such that

$A=S^{-1}CS$

Thus, matrix C is not similar to matrix A.

Step-by-step solution

Definition of Similar Matrix

Let A and B are two square matrices then matrix A is similar to matrix B if there exists an invertible matrix P such that

$B=P^{-1}AP$$B=P^{-1}AP$

Given the statement

Here it is given, that matrix A is similar to matrix B, and matrix B is similar to matrix C, then there exist two invertible matrices P and Q such that

$$\begin{gathered} B=P^{-1}AP \\ C=Q^{-1}BQ \end{gathered}$$

Making assumptions 

Let,

$A=\left[\begin{array}{cc}2& 1\\ 1& 2\end{array}\right],B=\left[\begin{array}{cc}-1& -8\\ 1& 5\end{array}\right],C=\left[\begin{array}{cc}-5& -28\\ 1& 5\end{array}\right]$

Since matrix A is similar to matrix B then there exists an invertible matrix $P=\left[\begin{array}{cc}1& 3\\ 0& 1\end{array}\right]$ such that

$B=P^{-1}AP$

Also, since matrix B is similar to matrix C, then there exists an invertible matrix $Q=\left[\begin{array}{cc}1& 4\\ 0& 1\end{array}\right]$ such that

$C=Q^{-1}BQ$

To find whether the matrix C is similar to matrix A or not

We assumed that

$A=\left[\begin{array}{cc}2& 1\\ 1& 2\end{array}\right]andC=\left[\begin{array}{cc}-5& -28\\ 1& 5\end{array}\right]$

Let us suppose that matrix C is similar to matrix A. Then there exists an invertible matrix $S=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$ such that

$A=S^{-1}CS$

$$\begin{gathered} \Rightarrow CS=SA \\ \Rightarrow \left[\begin{array}{cc}-5& -28\\ 1& 5\end{array}\right]\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\left[\begin{array}{cc}2& 1\\ 1& 2\end{array}\right] \\ \Rightarrow \left[\begin{array}{cc}-5a-28c& -5b-28d\\ a+5c& b+5d\end{array}\right]=\left[\begin{array}{cc}2a+b& a+2b\\ 2c+d& c+2d\end{array}\right] \end{gathered}$$

$$\begin{gathered} \Rightarrow -5a-28c=2a+b \\ -5b-28d=a+2b \\ a+5c=2c+d \\ b+5d=c+2d \end{gathered}$$

Writing the above equation in matrix form

The above equation in matrix form can be written as-

$\left[\begin{array}{cccc}1& 0& 3& -1\\ 0& 1& -1& 3\\ 1& 7& 0& 28\\ 7& 1& 28& 0\end{array}\right]\left[\begin{array}{c}a\\ b\\ c\\ d\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\\ 0\end{array}\right]$

Applying $R_3\to R_3-R_1,\text{and}{R}_4\to 7R_1-R_4.$

$\left[\begin{array}{cccc}1& 0& 3& -1\\ 0& 1& -1& 3\\ 0& 7& -3& -29\\ 0& -1& -7& -7\end{array}\right]\left[\begin{array}{c}a\\ b\\ c\\ d\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\\ 0\end{array}\right]$

Applying $R_3\to R_3-7R_2,\text{and}{R}_4\to R_2+R_4.$

$\left[\begin{array}{cccc}1& 0& 3& -1\\ 0& 1& -1& 3\\ 0& 0& 4& -50\\ 0& 0& -8& -4\end{array}\right]\left[\begin{array}{c}a\\ b\\ c\\ d\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\\ 0\end{array}\right]$

Applying$R_4\to 2R_3+R_4.$

$\left[\begin{array}{cccc}1& 0& 3& -1\\ 0& 1& -1& 3\\ 0& 0& 4& -50\\ 0& 0& 0& -104\end{array}\right]\left[\begin{array}{c}a\\ b\\ c\\ d\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\\ 0\end{array}\right]$

To find the value of a, b, c, d.

From the above matrix, we have

$$\begin{gathered} a+3c=0 \\ \text{b - c + 3d = 0} \\ 4c-50d=0 \\ -104d=0 \end{gathered}$$

$\Rightarrow d=0,a=0,=0,c=0$

Therefore, matrix $S=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$ which is not invertible.

Thus, matrix C is not similar to matrix A.

Final Answer

Thus, if a matrix A is similar to a matrix B, and the matrix B is similar to a matrix C, then there does not exist any invertible matrix S such that

$A=S^{-1}CS$

Hence, matrix C is not similar to matrix A.