It is not true.
Let $(R,m,k)$ be a Cohen-Macaulay local ring of dimension $d$ and $N$ is the canonical module of $R$. Let $M$ be a CohenMacaulay R-module of dimension $t$. Then, by Theorem.3.3.10 of Bruns-Herzog $Ext^{d-t}(M,N)$ is a Cohen-Macaulay R-module of dimension $t$.
So if $d-t\gt 0$, then $\dim Ext^{d-t}(M,N) =t =\dim M $

It is not true.
Let $(R,m,k)$ be a Cohen-Macaulay local ring of dimension $d$ and $N$ be the canonical module of $R$. Let $M$ be a Cohen-Macaulay $R$-module of dimension $t$. Then, by Theorem 3.3.10 of Bruns-Herzog $\text{Ext}^{d-t}(M,N)$ is a Cohen-Macaulay $R$-module of dimension $t$.
So if $d-t\gt 0$, then $\dim \text{Ext}^{d-t}(M,N) =t =\dim M $

Let $(R,m,k)$ be a Cohen-Macaulay local ring of dimension $d$ and $N$ is the canonical module of $R$. Let $M$ be a CohenMacaulay R-module of dimension $t$. Then, by Theorem.3.3.10 of Bruns-Herzog $Ext^{d-t}(M,N)$ is a Cohen-Macaulay R-module of dimension $t$.
So if $d-t\gt 0$, then $\dim Ext^{d-t}(M,N) =t =\dim M $

It is not true.
Let $(R,m,k)$ be a Cohen-Macaulay local ring of dimension $d$ and $N$ is the canonical module of $R$. Let $M$ be a CohenMacaulay R-module of dimension $t$. Then, by Theorem.3.3.10 of Bruns-Herzog $Ext^{d-t}(M,N)$ is a Cohen-Macaulay R-module of dimension $t$.
So if $d-t\gt 0$, then $\dim Ext^{d-t}(M,N) =t =\dim M $